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First published on Saturday, Jan 11, 2025 and last modified on Saturday, Jan 11, 2025
Mathedu
Singular Values, SVD
In that test, you will use the fact that, for any \( 2\times 2\) matrix \( A\) with real elements, the matrix \( A^TA\) is symmetric and thus, as you have seen in the section 11 test, it is diagonalizable in an orthonormal basis, to define the singular values of \( A\) .
For all the test, assume that \( A\) is a \( 2\times 2\) matrix with real elements.
Assume that \( X\) is a column vector with two real elements.
Prove that \( X^TA^TAX\ge 0\) .
Consider the linear mapping \( f:\mathbb{P}\rightarrow\mathbb{P}\) with matrix \( A\) in the canonical basis, and the vector \( \overrightarrow{u}\in\mathbb{P}\) with column vector of coordinates \( X\) in the canonical basis.
Consider a unitary matrix \( U\) and the diagonal matrix \( D\) such that \( A^TA=UDU^T\) .
Consider the vector \( \overrightarrow{u}\in\mathbb{P}\) with column vector of coordinates \( X\) in the canonical basis.
Then the coordinates of \( \overrightarrow{u}\) in the basis made of the vectors with column vectors of coordinates in the canonical basis the columns of \( A\) are \( Y=U^{-1}X=U^TX\) .
Prove that \( Y^TDY=X^TA^TAX\ge 0\) .
Use the fact that \( U\) is unitary to express \( D\) as a function of \( A\) and \( U\) .
Use the previous question to prove that the eigenvalues of \( A^TA\) are all positive or zero.
Use the fact that the diagonal elements \( \lambda_1\) and \( \lambda_2\) of matrix \( D\) considered in the previous question are the eigenvalues of \( A^TA\) , and consider the vectors \( Y\) with elements \( 1\) and \( 0\) and \( 0\) and \( 1\) .
Consequently, as \( \lambda_1\) and \( \lambda_2\) are the eigenvalues of \( A^TA\) , the eigenvalues of \( A^TA\) are positive or zero.
The singular values of \( A\) are defined as the square roots of the eigenvalues of \( A^TA\) .
Assume that the singular values of \( A\) are non zero, that is that they are positive.
\begin{myList}{numbered} \item Prove that \( A^TA\) is invertible.
Consider the unitary matrix \( U\) and the diagonal matrix \( D\) such that \( A^TA=UDU^T\) .
\item Prove by contradiction that \( A\) is invertible.
Use the fact that a matrix \( M\) is not invertible if and only if there exits a non zero column vector \( X\) such that \( MX=O_{21}\) , the null column vector. \end{myList}
Assume that \( 0\) is a singular value of \( A\) , so that \( 0\) is an eigenvalue of \( A^TA\) , and consider a column vector \( X\) such that \( A^TAX=O_{21}\) , the null column vector.
Prove that \( AX=O_{21}\) , so that \( 0\) is an eigenvalue of \( A\) .
Use the fact that \( X^TA^TAX\) is the square of the norm of the vector \( f(\overrightarrow{u})\) , where \( f\) is the linear mapping with matrix \( A\) in the canonical basis, and \( \overrightarrow{u}\in\mathbb{P}\) is the vector with column vector of coordinates \( X\) in the canonical basis.
Prove by contradiction that if \( 0\) is a singular value of \( A\) , then \( A\) is not invertible.
Consider a non zero column vector \( X\) such that \( A^TAX=O_{21}\) , the null column vector.
We have proved the following facts.
The singular values of a \( 2\times 2\) matrix with real elements are all positive or zero.
If the eigenvalues of a matrix are positive, then that matrix is invertible.
A matrix having \( 0\) as singular value is not invertible.
We have thus proved that a matrix is invertible if and only if its singular values are positive.