Given \(A \in \mathbb{R}^{n \times n}\) we say \(\lambda \in \mathbb{C}\) is an eigenvalue of \(A\) and \(v \in \mathbb{C}^{n} \setminus \{0\}\) is an eigenvector of \(A\) associated with the eigenvalue \(\lambda\) when the following holds: \[ Av = \lambda v \]
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Given \(A \in \mathbb{R}^{n \times n}\) we say \(\lambda \in \mathbb{C}\) is an eigenvalue of \(A\) and \(v \in \mathbb{C}^{n} \setminus \{0\}\) is an eigenvector of \(A\) associated with the eigenvalue \(\lambda\) when the following holds: \[ Av = \lambda v \]
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| Text | Given \(A \in \mathbb{R}^{n \times n}\) we say \(\lambda \in \mathbb{C}\) is an {{c2::<b>eigenvalue</b>}} of \(A\) and \(v \in \mathbb{C}^{n} \setminus \{0\}\) is an {{c2::<b>eigenvector</b> of \(A\) associated with the eigenvalue \(\lambda\)}} when the following holds: \[{{c1:: Av = \lambda v }}\] |
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The eigenvectors of \(A^{-1}\) are the same as those of \(A\).
Back
The eigenvectors of \(A^{-1}\) are the same as those of \(A\).
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| Text | The eigenvectors of \(A^{-1}\) are {{c1::the same}} as those of \(A\). |
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The matrix \(A^k\) has EW-EW pair \(\lambda^k\) and \(v\) if \(A\) has \(\lambda, v\) as an EW-EV pair.
Back
The matrix \(A^k\) has EW-EW pair \(\lambda^k\) and \(v\) if \(A\) has \(\lambda, v\) as an EW-EV pair.
Intuitively, \(A\) on \(v\) scales it by \(\lambda\). Then scaling that already scaled \(\lambda v\) by \(A\) again gives us \(\lambda^2 v\), etc...
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| Text | The matrix \(A^k\) has EW-EW pair {{c1::\(\lambda^k\) and \(v\)}} if \(A\) has \(\lambda, v\) as an EW-EV pair. | |
| Extra | <i>Intuitively</i>, \(A\) on \(v\) scales it by \(\lambda\). Then scaling that already scaled \(\lambda v\) by \(A\) again gives us \(\lambda^2 v\), etc... |
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Let \(A\) be an invertible matrix. If \(\lambda\) and \(v\) are an EW-EV pair, then {{c1:: \(\frac{1}{\lambda}\) and \(v\)}} are an EW-EV pair of the matrix \(A^{-1}\). Proof Included
Back
Let \(A\) be an invertible matrix. If \(\lambda\) and \(v\) are an EW-EV pair, then {{c1:: \(\frac{1}{\lambda}\) and \(v\)}} are an EW-EV pair of the matrix \(A^{-1}\). Proof Included
Proof: We have \(Av = \lambda v\) thus \(A^{-1}Av = \lambda A^{-1}v\) thus \(\frac{1}{\lambda} v = \frac{1}{\lambda} \lambda A^{-1}v\) and we get \(A^{-1}v = \frac{1}{\lambda} v\).
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| Text | Let \(A\) be an invertible matrix. If \(\lambda\) and \(v\) are an EW-EV pair, then {{c1:: \(\frac{1}{\lambda}\) and \(v\)}} are an EW-EV pair of the matrix \(A^{-1}\). <i>Proof Included</i> | |
| Extra | <div><b>Proof:</b> We have \(Av = \lambda v\) thus \(A^{-1}Av = \lambda A^{-1}v\) thus \(\frac{1}{\lambda} v = \frac{1}{\lambda} \lambda A^{-1}v\) and we get \(A^{-1}v = \frac{1}{\lambda} v\).</div> |
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\[ P(z) = (-1)^n \det(A - z I) = \det(z I - A) = (z - \lambda_1)(z - \lambda_2) \dots (z - \lambda_n) \]
The polynomial \(P(z)\) is called the characteristic polynomial of the matrix \(A\).
Back
\[ P(z) = (-1)^n \det(A - z I) = \det(z I - A) = (z - \lambda_1)(z - \lambda_2) \dots (z - \lambda_n) \]
The polynomial \(P(z)\) is called the characteristic polynomial of the matrix \(A\).
The eigenvalues \(\lambda_1, \dots, \lambda_n\) as they show up in the polynomial are not all distinct in general.
The number of times an eigenvalue shows up is called the algebraic multiplicity of the eigenvalue.
Field-by-field Comparison
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| Text | <div>\[ P(z) = {{c22:: (-1)^n \det(A - z I) = \det(z I - A) = (z - \lambda_1)(z - \lambda_2) \dots (z - \lambda_n) }}\]</div><div>The polynomial \(P(z)\) is called {{c1::the characteristic polynomial of the matrix \(A\)}}.</div> | |
| Extra | <div>The eigenvalues \(\lambda_1, \dots, \lambda_n\) as they show up in the polynomial are <b>not all distinct</b> in general.</div><div>The number of times an eigenvalue shows up is called the <b>algebraic multiplicity</b> of the eigenvalue.</div> |
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Let \(Q\) be an orthogonal matrix. If \(\lambda \in \mathbb{C}\) is an eigenvalue of \(Q\) then \(|\lambda| =\)\(1\).
Back
Let \(Q\) be an orthogonal matrix. If \(\lambda \in \mathbb{C}\) is an eigenvalue of \(Q\) then \(|\lambda| =\)\(1\).
The possible values for \(\lambda\) are then \(1, -1\) and all conjugate complex values with modulus \(1\) for example \(i, -i\).
This makes sense as \(Q\) orthogonal only turns and doesn't scale.
This makes sense as \(Q\) orthogonal only turns and doesn't scale.
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| Text | <div>Let \(Q\) be an orthogonal matrix. If \(\lambda \in \mathbb{C}\) is an eigenvalue of \(Q\) then \(|\lambda| =\){{c1::\(1\)}}.</div> | |
| Extra | The possible values for \(\lambda\) are then \(1, -1\) and all conjugate complex values with modulus \(1\) for example \(i, -i\).<br><br>This makes sense as \(Q\) orthogonal only turns and doesn't scale. |
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The eigenvalues of \(AB\) and \(BA\) are not correlated.
Back
The eigenvalues of \(AB\) and \(BA\) are not correlated.
Field-by-field Comparison
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| Text | The eigenvalues of \(AB\) and \(BA\) are {{c1::not correlated}}. |
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What is special about the characteristic polynomial?
Back
What is special about the characteristic polynomial?
The characteristic polynomial is always monic.
The polynomial \(\det(A - zI)\) has a leading \((-1)\) if the degree is odd. Therefore working with the characteristic one is easier.
The polynomial \(\det(A - zI)\) has a leading \((-1)\) if the degree is odd. Therefore working with the characteristic one is easier.
Field-by-field Comparison
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| Front | What is special about the characteristic polynomial? | |
| Back | The characteristic polynomial is always <b>monic</b>.<br><br>The polynomial \(\det(A - zI)\) has a leading \((-1)\) if the degree is odd. Therefore working with the characteristic one is easier. |
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The eigenvectors of an eigenvalue are those and exactly those vectors \(v \neq 0\) in \(v \in N(A - \lambda I)\) .
Back
The eigenvectors of an eigenvalue are those and exactly those vectors \(v \neq 0\) in \(v \in N(A - \lambda I)\) .
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| Text | The <b>eigenvectors</b> of an eigenvalue are <b>those and exactly those</b> vectors {{c1::\(v \neq 0\)}} in {{c1::\(v \in N(A - \lambda I)\) :: subspace}}. |
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Let \(A \in \mathbb{R}^{n \times n}\). If \((\lambda, v)\) is an eigenvalue, eigenvector pair, then {{c1::\((\overline{\lambda}, \overline{v})\) }} is an eigenvalue, eigenvector pair.
Back
Let \(A \in \mathbb{R}^{n \times n}\). If \((\lambda, v)\) is an eigenvalue, eigenvector pair, then {{c1::\((\overline{\lambda}, \overline{v})\) }} is an eigenvalue, eigenvector pair.
The conjugate is always also an EW, EV pair.
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| Text | Let \(A \in \mathbb{R}^{n \times n}\). If \((\lambda, v)\) is an eigenvalue, eigenvector pair, then {{c1::\((\overline{\lambda}, \overline{v})\) }} is an eigenvalue, eigenvector pair. | |
| Extra | The conjugate is always also an EW, EV pair. |
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The eigenvalues of \(A^{-1}\) are \(1/\lambda_i\) if \(\lambda_i\)'s are the eigenvalues of \(A\).
Back
The eigenvalues of \(A^{-1}\) are \(1/\lambda_i\) if \(\lambda_i\)'s are the eigenvalues of \(A\).
Field-by-field Comparison
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| Text | The eigenvalues of \(A^{-1}\) are {{c1:: \(1/\lambda_i\) }} if \(\lambda_i\)'s are the eigenvalues of \(A\). |
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All eigenvalues are exactly the roots of the polynomial \(\det(A - \lambda I)\) .
Back
All eigenvalues are exactly the roots of the polynomial \(\det(A - \lambda I)\) .
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| Text | <div>All eigenvalues are {{c1::exactly the roots of the polynomial \(\det(A - \lambda I)\) :: in terms of polynomial }}.</div> |
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If \(Av = \lambda v\) and \(\lambda\) and \(v\) are an EW, EV pair we know \(v \neq 0\) .
Back
If \(Av = \lambda v\) and \(\lambda\) and \(v\) are an EW, EV pair we know \(v \neq 0\) .
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| Text | If \(Av = \lambda v\) and \(\lambda\) and \(v\) are an EW, EV pair we know {{c1::\(v \neq 0\) :: property of \(v\)}}. |
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The eigenvalues of \(A + B\) are not the eigenvalues of \(A\) plus those of \(B\).
Back
The eigenvalues of \(A + B\) are not the eigenvalues of \(A\) plus those of \(B\).
They aren't correlated.
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| Text | The eigenvalues of \(A + B\) are {{c1::<b>not</b> the eigenvalues of \(A\) plus those of \(B\)}}. | |
| Extra | They aren't correlated. |
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Let \(A\) and \(v_1, \dots, v_k \in \mathbb{R}^n\) be the EVs of \(\lambda_1, \dots, \lambda_k \in \mathbb{R}^n\).
If the \(\lambda_1, \dots, \lambda_k\)s are all distinct then the EVs \(v_1, \dots, v_k\) are all linearly independent.
Back
Let \(A\) and \(v_1, \dots, v_k \in \mathbb{R}^n\) be the EVs of \(\lambda_1, \dots, \lambda_k \in \mathbb{R}^n\).
If the \(\lambda_1, \dots, \lambda_k\)s are all distinct then the EVs \(v_1, \dots, v_k\) are all linearly independent.
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| Text | <div>Let \(A\) and \(v_1, \dots, v_k \in \mathbb{R}^n\) be the EVs of \(\lambda_1, \dots, \lambda_k \in \mathbb{R}^n\).</div><div>If the \(\lambda_1, \dots, \lambda_k\)s are all distinct then {{c1::the EVs \(v_1, \dots, v_k\) are all linearly independent}}.</div> |
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Let \(A\) with \(n\) distinct real eigenvalues (meaning that the zeros of \(\det(A- \lambda I)\) as described in Corollary 8.1.3 are all distinct, algebraic multiplicity \(1\)) , then {{c2::there is a basis of \(\mathbb{R}^n\) made up of EVs of \(A\)}}.
Back
Let \(A\) with \(n\) distinct real eigenvalues (meaning that the zeros of \(\det(A- \lambda I)\) as described in Corollary 8.1.3 are all distinct, algebraic multiplicity \(1\)) , then {{c2::there is a basis of \(\mathbb{R}^n\) made up of EVs of \(A\)}}.
We also call this the Eigenbasis or a complete set of real EVs, which will come in handy later for Diagonalisation.
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| Text | <div>Let \(A\) with \(n\) {{c1::distinct real eigenvalues (meaning that the zeros of \(\det(A- \lambda I)\) as described in Corollary 8.1.3 are all distinct, algebraic multiplicity \(1\)) :: property and in terms of algebraic }}, then {{c2::there is a basis of \(\mathbb{R}^n\) made up of EVs of \(A\)}}.</div> | |
| Extra | <div>We also call this the <b>Eigenbasis</b> or a <b>complete set of real EVs</b>, which will come in handy later for Diagonalisation.</div> |
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Real Antisymmetric matrices always have imaginary (or zero) eigenvalues.
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Real Antisymmetric matrices always have imaginary (or zero) eigenvalues.
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| Text | <div>Real Antisymmetric matrices always have {{c1::imaginary (or zero) eigenvalues}}.</div> |
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A real valued matrix \(A \in \mathbb{R}^{n \times n}\) has real or complex valued eigenvalues.
Back
A real valued matrix \(A \in \mathbb{R}^{n \times n}\) has real or complex valued eigenvalues.
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| Text | <div>A real valued matrix \(A \in \mathbb{R}^{n \times n}\) has {{c1::real <b>or </b>complex}} valued eigenvalues.</div> |
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A real EV always has a real EW associated with it.
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A real EV always has a real EW associated with it.
Field-by-field Comparison
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| Text | {{c1::A real EV}} always has {{c2::a real EW}} associated with it. |
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For an eigenvalue \(\lambda\) of \(M\), \(\lambda + c\) is a real eigenvalue of the matrix \(M + cI\).
Back
For an eigenvalue \(\lambda\) of \(M\), \(\lambda + c\) is a real eigenvalue of the matrix \(M + cI\).
Intuitively this makes sense as by adding \(cI\) were increasing the values on the diagonal, meaning we have to increase the value of \(\lambda\) by the same amount so it makes \(0\) again.
Field-by-field Comparison
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| Text | For an eigenvalue \(\lambda\) of \(M\), {{c1::\(\lambda + c\)}} is a real eigenvalue of the matrix {{c2::\(M + cI\)}}. | |
| Extra | Intuitively this makes sense as by adding \(cI\) were increasing the values on the diagonal, meaning we have to increase the value of \(\lambda\) by the same amount so it makes \(0\) again. |
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The eigenvectors of \(A\) are not the same as those of \(A^\top\).
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The eigenvectors of \(A\) are not the same as those of \(A^\top\).
Field-by-field Comparison
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| Text | The eigenvectors of \(A\) are {{c1::<b>not the same</b>}} as those of \(A^\top\). |
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All the eigenvectors for \(\lambda_i\) are the vectors \(v \neq 0\), \(v \in N(A - \lambda_i I)\), in the nullspace .
Back
All the eigenvectors for \(\lambda_i\) are the vectors \(v \neq 0\), \(v \in N(A - \lambda_i I)\), in the nullspace .
Field-by-field Comparison
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| Text | <div>All the eigenvectors for \(\lambda_i\) are {{c1::the vectors \(v \neq 0\), \(v \in N(A - \lambda_i I)\), in the nullspace :: subspace}}.</div> |
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Let \(A \in \mathbb{R}^{n \times n}\) and \(\lambda_1, \dots, \lambda_n\) its \(n\) eigenvalues. Then \[ \det(A) = {{c1:: \prod_{i = 1}^n \lambda_i :: \text{in terms of EWs} }}\]
Back
Let \(A \in \mathbb{R}^{n \times n}\) and \(\lambda_1, \dots, \lambda_n\) its \(n\) eigenvalues. Then \[ \det(A) = {{c1:: \prod_{i = 1}^n \lambda_i :: \text{in terms of EWs} }}\]
Be careful to include each eigenvalue as often as their algebraic multiplicity in these sums/products. You can use this to double check calculations.
Intuition: The eigenvalues describe how much each eigenvector is scaled. Thus by multiplying the scaling of each dimension, we can figure out the volume of the unit cube which is the determinant.
Field-by-field Comparison
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| Text | Let \(A \in \mathbb{R}^{n \times n}\) and \(\lambda_1, \dots, \lambda_n\) its \(n\) eigenvalues. Then \[ \det(A) = {{c1:: \prod_{i = 1}^n \lambda_i :: \text{in terms of EWs} }}\] | |
| Extra | <div><strong>Be careful</strong> to include each eigenvalue as often as their <em>algebraic multiplicity</em> in these sums/products. You can use this to double check calculations.</div><div><br></div><div><i>Intuition:</i> The eigenvalues describe how much each eigenvector is scaled. Thus by multiplying the scaling of each dimension, we can figure out the <i>volume of the unit cube</i> which is the determinant.</div> |
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\(\text{Tr}(AB) = {{c1:: \text{Tr}(BA) }}\) and \(\text{Tr}(ABC) = {{c1:: \text{Tr}(CBA) = \text{Tr}(BAC) }}\)
Back
\(\text{Tr}(AB) = {{c1:: \text{Tr}(BA) }}\) and \(\text{Tr}(ABC) = {{c1:: \text{Tr}(CBA) = \text{Tr}(BAC) }}\)
The trace is commutative.
Field-by-field Comparison
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| Text | \(\text{Tr}(AB) = {{c1:: \text{Tr}(BA) }}\) and \(\text{Tr}(ABC) = {{c1:: \text{Tr}(CBA) = \text{Tr}(BAC) }}\) | |
| Extra | The trace is commutative. |
Note 25: ETH::LinAlg
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Give an example of a matrix with complex valued EWs:
Back
Give an example of a matrix with complex valued EWs:
Eigenvalues of the \(90^\circ\) degree counterclockwise rotation matrix \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\).
The solutions to \(0 = \det(A - \lambda I) = -\lambda \cdot -\lambda - 1 \cdot (-1) = \lambda^2 + 1\) which are \(\lambda_1 = i\) and \(\lambda_2 = -i\). The eigenvectors are given by \(v_1 = \begin{pmatrix} i \\ 1 \end{pmatrix}\) \(v_2 = \begin{pmatrix} -i \\ 1 \end{pmatrix}\).
This makes sense because the only vector staying on it's axis in a 2d rotation of a plane by \(90^\circ\) is the vector pointing straight up, out from the plane.
The solutions to \(0 = \det(A - \lambda I) = -\lambda \cdot -\lambda - 1 \cdot (-1) = \lambda^2 + 1\) which are \(\lambda_1 = i\) and \(\lambda_2 = -i\). The eigenvectors are given by \(v_1 = \begin{pmatrix} i \\ 1 \end{pmatrix}\) \(v_2 = \begin{pmatrix} -i \\ 1 \end{pmatrix}\).
This makes sense because the only vector staying on it's axis in a 2d rotation of a plane by \(90^\circ\) is the vector pointing straight up, out from the plane.
Field-by-field Comparison
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| Front | Give an example of a matrix with complex valued EWs: | |
| Back | Eigenvalues of the \(90^\circ\) degree counterclockwise rotation matrix \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\).<br><br>The solutions to \(0 = \det(A - \lambda I) = -\lambda \cdot -\lambda - 1 \cdot (-1) = \lambda^2 + 1\) which are \(\lambda_1 = i\) and \(\lambda_2 = -i\). The eigenvectors are given by \(v_1 = \begin{pmatrix} i \\ 1 \end{pmatrix}\) \(v_2 = \begin{pmatrix} -i \\ 1 \end{pmatrix}\).<br><br>This makes sense because the only vector staying on it's axis in a 2d rotation of a plane by \(90^\circ\) is the vector pointing straight up, out from the plane. |
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Let \(A \in \mathbb{R}^{n \times n}\) and \(\lambda_1, \dots, \lambda_n\) its \(n\) eigenvalues. Then \[ \text{Tr}(A) = {{c1:: \sum_{i = 1}^n \lambda_i :: \text{in terms of EWs} }}\]
Back
Let \(A \in \mathbb{R}^{n \times n}\) and \(\lambda_1, \dots, \lambda_n\) its \(n\) eigenvalues. Then \[ \text{Tr}(A) = {{c1:: \sum_{i = 1}^n \lambda_i :: \text{in terms of EWs} }}\]
Quite surprising, since the determinant and trace are \(\in \mathbb{R}\) where the eigenvalues in general must not be?
It holds because complex eigenvalues \(z_1, z_2\) always show up in pairs: \(z_1 = \overline{z_2}\). And because \(z_1 \cdot \overline{z_1} = a^2 + b^2\) and \(z_1 + \overline{z_1} = 2a\).
It holds because complex eigenvalues \(z_1, z_2\) always show up in pairs: \(z_1 = \overline{z_2}\). And because \(z_1 \cdot \overline{z_1} = a^2 + b^2\) and \(z_1 + \overline{z_1} = 2a\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | Let \(A \in \mathbb{R}^{n \times n}\) and \(\lambda_1, \dots, \lambda_n\) its \(n\) eigenvalues. Then \[ \text{Tr}(A) = {{c1:: \sum_{i = 1}^n \lambda_i :: \text{in terms of EWs} }}\] | |
| Extra | Quite surprising, since the determinant and trace are \(\in \mathbb{R}\) where the eigenvalues in general must not be?<br>It holds because complex eigenvalues \(z_1, z_2\) always show up in pairs: \(z_1 = \overline{z_2}\). And because \(z_1 \cdot \overline{z_1} = a^2 + b^2\) and \(z_1 + \overline{z_1} = 2a\). |
Note 27: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID:
added
Note Type: Horvath Classic
GUID:
lK,rOhy|Tw
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Front
If we have \(A\) with eigenvalues \(0, 1, 2\) then \(I + A^2\) has eigenvalues?
Back
If we have \(A\) with eigenvalues \(0, 1, 2\) then \(I + A^2\) has eigenvalues?
We have \(\lambda^2\) eigenvalues of \(A^2\) by lemma script. Thus \(0, 1, 4\) are EWs.
Then \(1 + \lambda^2\) are the eigenvalues of \(I + A^2\) thus \(1, 2, 5\) are the EWs.
Then \(1 + \lambda^2\) are the eigenvalues of \(I + A^2\) thus \(1, 2, 5\) are the EWs.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | If we have \(A\) with eigenvalues \(0, 1, 2\) then \(I + A^2\) has eigenvalues? | |
| Back | We have \(\lambda^2\) eigenvalues of \(A^2\) by lemma script. Thus \(0, 1, 4\) are EWs.<br>Then \(1 + \lambda^2\) are the eigenvalues of \(I + A^2\) thus \(1, 2, 5\) are the EWs. |
Note 28: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Classic
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Note Type: Horvath Classic
GUID:
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Front
How do we get from \(\det(A - zI)\) to \(\det(zI - A)\)?
Back
How do we get from \(\det(A - zI)\) to \(\det(zI - A)\)?
We can transform \(\det(A - zI) =\) \((-1)^n \det((-1)(A - zI))\) \(= (-1)^n \det(zI - A)\) because \(\det(\lambda A) = \lambda^n \det(A)\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | How do we get from \(\det(A - zI)\) to \(\det(zI - A)\)? | |
| Back | We can transform \(\det(A - zI) =\) \((-1)^n \det((-1)(A - zI))\) \(= (-1)^n \det(zI - A)\) because \(\det(\lambda A) = \lambda^n \det(A)\). |
Note 29: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Front
The eigenvalues of \(A\) are the same ones as those of \(A^\top\).
Back
The eigenvalues of \(A\) are the same ones as those of \(A^\top\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | The {{c1::eigenvalues}} of \(A\) are the same ones as those of \(A^\top\). |
Note 30: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Note Type: Horvath Cloze
GUID:
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Front
A vector \(v \in \mathbb{R}^n \setminus \{0\}\) is an eigenvector associated with the eigenvalue \(\lambda\) if and only if \(v \in N(A - \lambda I)\).
Back
A vector \(v \in \mathbb{R}^n \setminus \{0\}\) is an eigenvector associated with the eigenvalue \(\lambda\) if and only if \(v \in N(A - \lambda I)\).
Field-by-field Comparison
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|---|---|---|
| Text | A vector \(v \in \mathbb{R}^n \setminus \{0\}\) is {{c1::an eigenvector associated with the eigenvalue \(\lambda\)}} if and only if {{c2::\(v \in N(A - \lambda I)\)}}. |
Note 31: ETH::LinAlg
Deck: ETH::LinAlg
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Note Type: Horvath Cloze
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Front
An EW can have many EVs associated with it.
Back
An EW can have many EVs associated with it.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | An EW can have {{c1::many :: number}} EVs associated with it. |
Note 32: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Note Type: Horvath Cloze
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Front
Let \(A \in \mathbb{R}^{n \times n}\).
\(\lambda \in \mathbb{R}\) is a real eigenvalue of \(A\) if and only if \(\det(A - \lambda I) = 0\).
Back
Let \(A \in \mathbb{R}^{n \times n}\).
\(\lambda \in \mathbb{R}\) is a real eigenvalue of \(A\) if and only if \(\det(A - \lambda I) = 0\).
Field-by-field Comparison
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|---|---|---|
| Text | <div>Let \(A \in \mathbb{R}^{n \times n}\).</div><div>\(\lambda \in \mathbb{R}\) is a {{c1::real eigenvalue}} of \(A\) if and only if {{c2::\(\det(A - \lambda I) = 0\)}}. </div> |
Note 33: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Note Type: Horvath Cloze
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Front
Every matrix \(A \in \mathbb{R}^{n \times n}\) has an eigenvalue (perhaps complex-valued) .
Back
Every matrix \(A \in \mathbb{R}^{n \times n}\) has an eigenvalue (perhaps complex-valued) .
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | Every matrix \(A \in \mathbb{R}^{n \times n}\) has {{c1::an eigenvalue (perhaps <i>complex</i>-valued) :: due to fundamental theorem of algebra}}. |
Note 34: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Note Type: Horvath Cloze
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Front
The eigenvalues of \(AB\) and \(BA\) are the same.
Back
The eigenvalues of \(AB\) and \(BA\) are the same.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | The eigenvalues of \(AB\) and \(BA\) are {{c1::the same}}. |
Note 35: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID:
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Note Type: Horvath Cloze
GUID:
pk}0Q0,75,
Previous
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Front
The trace is commutative
Back
The trace is commutative
This makes sense as addition is element-wise.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | The trace is {{c1::commutative}} | |
| Extra | This makes sense as <b>addition</b> is <b>element-wise</b>. |
Note 36: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Note Type: Horvath Cloze
GUID:
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Front
\(A\) has an EW \(0\) \(\Longleftrightarrow\)\(A\) is not invertible Proof Included
Back
\(A\) has an EW \(0\) \(\Longleftrightarrow\)\(A\) is not invertible Proof Included
We know the EVs are in the \(N(A - \lambda I)\) because they solve the formula \(Av = \lambda v \implies Av - \lambda v \)\(= 0 \implies (A - \lambda I)v = 0\). Thus \(v\) is in the nullspace of \((A - \lambda I)\).
If \(A\) is not invertible, then it will have an EV of \(0\), which means that there is an EV solving \((A - 0 \cdot I)v = 0 \implies Av = 0\). Thus \(v \neq 0\) is in the nullspace of \(A\), i.e. the nullspace is not empty.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | {{c1::\(A\) has an EW \(0\)}} \(\Longleftrightarrow\){{c2::\(A\) is not invertible}}<i> Proof Included</i> | |
| Extra | <div>We know the EVs are in the \(N(A - \lambda I)\) because they solve the formula \(Av = \lambda v \implies Av - \lambda v \)\(= 0 \implies (A - \lambda I)v = 0\). Thus \(v\) is in the nullspace of \((A - \lambda I)\).</div><div><br></div><div>If \(A\) is not invertible, then it will have an EV of \(0\), which means that there is an EV solving \((A - 0 \cdot I)v = 0 \implies Av = 0\). Thus \(v \neq 0\) is in the nullspace of \(A\), i.e. the nullspace is not empty.</div> |
Note 37: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID:
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Note Type: Horvath Cloze
GUID:
ww{@M/WDmP
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Front
\(A\) and \(A^\top\) share eigenvalues not eigenvectors .
Back
\(A\) and \(A^\top\) share eigenvalues not eigenvectors .
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <div>\(A\) and \(A^\top\) share {{c1::eigenvalues <b>not eigenvectors</b> :: EWs, EVs}}.</div> |
Note 38: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID:
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Note Type: Horvath Cloze
GUID:
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Front
Gaussian Elimination does not preserve EWs or EVs.
Back
Gaussian Elimination does not preserve EWs or EVs.
This means the EVs and EWs depend on the representation of the matrix, not on the subspaces they define.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <div>Gaussian Elimination does {{c1::<b>not</b>}} preserve EWs or EVs. </div> | |
| Extra | This means the EVs and EWs depend on the representation of the matrix, not on the subspaces they define. |
Note 39: ETH::LinAlg
Deck: ETH::LinAlg
Note Type: Horvath Cloze
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Note Type: Horvath Cloze
GUID:
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Previous
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Front
Real Symmetric matrices always have real eigenvalues.
Back
Real Symmetric matrices always have real eigenvalues.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <div>Real Symmetric matrices always have {{c1::real eigenvalues}}.</div> |