Anki Deck Changes

Note 1: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: Ny?NMfTURP

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Front

\(A^\top A\) has full rank when \(A\) has full column rank. Proof Included

Back

\(A^\top A\) has full rank when \(A\) has full column rank. Proof Included

If \(A^\top A x = 0\), then \(x^\top A^\top A x = ||Ax||^2 = 0\), so \(Ax = 0\). If A has full column rank \(N(A) = \{0\}\), thus \(x = 0\), proving \(A^\top A\) is invertible - full rank (as \(A^\top A \in \mathbb{R}^{n \times n}\)).

Field-by-field Comparison

Field	Before	After
Text		\(A^\top A\) has full rank when \(A\) has {{c1::full column rank}}. <i>Proof Included</i>
Extra		<div>If \(A^\top A x = 0\), then \(x^\top A^\top A x = \|\|Ax\|\|^2 = 0\), so \(Ax = 0\). If A has full column rank \(N(A) = \{0\}\), thus \(x = 0\), proving \(A^\top A\) is invertible - full rank (as \(A^\top A \in \mathbb{R}^{n \times n}\)).</div><div><br></div>

Tags: ETH::1._Semester::LinAlg::9._Diagonalisable_Matrices_and_the_SVD::1._Diagonalisation

Note 2: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: PDdrc~!V)N

added

Note did not exist

New Note

Front

\(AA^\top\) has full rank when \(A\) has full row rank.

Back

\(AA^\top\) has full rank when \(A\) has full row rank.

\(AA^\top x = 0\) implies \(x^\top AA^\top x = 0 \implies ||A^\top x||^2 = 0\) thus \(x \in N(A^\top)\). And for \(A\) full row rank, \(N(A^\top) = \{0\}\). Thus \(x = 0\) and \(AA^\top \) has full rank - invertible.

Field-by-field Comparison

Field	Before	After
Text		\(AA^\top\) has full rank when \(A\) has {{c1::full row rank}}.
Extra		\(AA^\top x = 0\) implies \(x^\top AA^\top x = 0 \implies \|\|A^\top x\|\|^2 = 0\) thus \(x \in N(A^\top)\). And for \(A\) full row rank, \(N(A^\top) = \{0\}\). Thus \(x = 0\) and \(AA^\top \) has full rank - invertible.

Tags: ETH::1._Semester::LinAlg::9._Diagonalisable_Matrices_and_the_SVD::1._Diagonalisation

Note 3: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Cloze
GUID: dmvZr9zGs8

added

Note did not exist

New Note

Front

A diagonal matrix has it's EWs on the diagonal.

Back

A diagonal matrix has it's EWs on the diagonal.

Field-by-field Comparison

Field	Before	After
Text		A diagonal matrix has it's EWs {{c1:: on the diagonal}}.

Tags: ETH::1._Semester::LinAlg::9._Diagonalisable_Matrices_and_the_SVD::1._Diagonalisation

Note 4: ETH::LinAlg

Deck: ETH::LinAlg
Note Type: Horvath Classic
GUID:

g|NuPS1+


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                        Front
                        
                            
  ETH::1._Semester::LinAlg::2._Matrices
  
  
    Give the definition of a 2x2 rotation matrix.
  



                        
                    
                    
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  ETH::1._Semester::LinAlg::2._Matrices
  
  
    Give the definition of a 2x2 rotation matrix.
    
    \[A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\]where \(\theta\) is the rotation angle.
  






                        
                    
                    
                
            

            

            
            
            
                Field-by-field Comparison
                
                    
                        
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                            Front
                            
                            Give the definition of a 2x2 rotation matrix.
                        
                        
                        
                            Back
                            
                            \[A = \begin{bmatrix} \cos \theta &amp; -\sin \theta \\ \sin \theta &amp; \cos \theta \end{bmatrix}\]where&nbsp;\(\theta\)&nbsp;is the rotation angle.
                        
                        
                    
                
            
            

            
            
            
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                    ETH::1._Semester::LinAlg::2._Matrices

Field	Before	After
Front		Give the definition of a 2x2 rotation matrix.
Back		\[A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\]where \(\theta\) is the rotation angle.


        
        
            Note 5: ETH::LinAlg
            
                Deck: ETH::LinAlg

                Note Type: Horvath Classic

                GUID: vMw:@b$pAy
            
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                        Front
                        
                            
  ETH::1._Semester::LinAlg::9._Diagonalisable_Matrices_and_the_SVD::1._Diagonalisation
  
  
    How to recover a matrix \(A\) from it's eigenvectors and eigenvalues (complete set)?
  



                        
                    
                    
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  ETH::1._Semester::LinAlg::9._Diagonalisable_Matrices_and_the_SVD::1._Diagonalisation
  
  
    How to recover a matrix \(A\) from it's eigenvectors and eigenvalues (complete set)?
    
    \(V\) the matrix with the eigenvectors of \(A\), orthogonal. Then we know \(AV = VD\) (\(Av_i = \lambda_i v_i\) in matrix form), with \(D = \Lambda\) the matrix with the eigenvalues on the diagonal.

Thus \(AVV^\top = VDV^\top \implies A = VDV^\top\) .
  






                        
                    
                    
                
            

            

            
            
            
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                            Front
                            
                            How to recover a matrix&nbsp;\(A\)&nbsp;from it's eigenvectors and eigenvalues (complete set)?
                        
                        
                        
                            Back
                            
                            \(V\)&nbsp;the matrix with the eigenvectors of&nbsp;\(A\), orthogonal. Then we know&nbsp;\(AV = VD\)&nbsp;(\(Av_i = \lambda_i v_i\)&nbsp;in matrix form), with&nbsp;\(D = \Lambda\)&nbsp;the matrix with the eigenvalues on the diagonal.<br><br>Thus&nbsp;\(AVV^\top = VDV^\top \implies A = VDV^\top\)&nbsp;.
                        
                        
                    
                
            
            

            
            
            
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                    ETH::1._Semester::LinAlg::9._Diagonalisable_Matrices_and_the_SVD::1._Diagonalisation

Field	Before	After
Front		How to recover a matrix \(A\) from it's eigenvectors and eigenvalues (complete set)?
Back		\(V\) the matrix with the eigenvectors of \(A\), orthogonal. Then we know \(AV = VD\) (\(Av_i = \lambda_i v_i\) in matrix form), with \(D = \Lambda\) the matrix with the eigenvalues on the diagonal.<br><br>Thus \(AVV^\top = VDV^\top \implies A = VDV^\top\) .