Is there an equivalent property of Similar Matrices for rectangular matrices?
$begingroup$
Say I have the following:
$Z = A^T B A$
where $Z$ is an $ntimes n$ matrix, $B$ is an $m times m$ matrix, and $A$ is a $m times n$ matrix, $n<m$.
Furthermore, the columns of $A$ are orthogonal, and $B$ is a positive definite matrix and $Z$ is symmetric tridiagonal.
This bears similarities to two similar $mtimes m$ matrices, $X$ and $Y$, requiring Y = $P^{-1}XP$ for some invertible matrix $P$.
Obviously in my case, $B$ and $Z$ don't have the same dimensions, and as $A$ is a truncated orthogonal matrix, it only has the property $A^TA=I$ (and not the other way round).
However, can any properties of similar matrices follow from this, such as the characteristic polynomial, determinant or eigenvalues?
linear-algebra abstract-algebra matrices
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
$endgroup$
add a comment |
$begingroup$
Say I have the following:
$Z = A^T B A$
where $Z$ is an $ntimes n$ matrix, $B$ is an $m times m$ matrix, and $A$ is a $m times n$ matrix, $n<m$.
Furthermore, the columns of $A$ are orthogonal, and $B$ is a positive definite matrix and $Z$ is symmetric tridiagonal.
This bears similarities to two similar $mtimes m$ matrices, $X$ and $Y$, requiring Y = $P^{-1}XP$ for some invertible matrix $P$.
Obviously in my case, $B$ and $Z$ don't have the same dimensions, and as $A$ is a truncated orthogonal matrix, it only has the property $A^TA=I$ (and not the other way round).
However, can any properties of similar matrices follow from this, such as the characteristic polynomial, determinant or eigenvalues?
linear-algebra abstract-algebra matrices
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
$endgroup$
add a comment |
$begingroup$
Say I have the following:
$Z = A^T B A$
where $Z$ is an $ntimes n$ matrix, $B$ is an $m times m$ matrix, and $A$ is a $m times n$ matrix, $n<m$.
Furthermore, the columns of $A$ are orthogonal, and $B$ is a positive definite matrix and $Z$ is symmetric tridiagonal.
This bears similarities to two similar $mtimes m$ matrices, $X$ and $Y$, requiring Y = $P^{-1}XP$ for some invertible matrix $P$.
Obviously in my case, $B$ and $Z$ don't have the same dimensions, and as $A$ is a truncated orthogonal matrix, it only has the property $A^TA=I$ (and not the other way round).
However, can any properties of similar matrices follow from this, such as the characteristic polynomial, determinant or eigenvalues?
linear-algebra abstract-algebra matrices
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
$endgroup$
Say I have the following:
$Z = A^T B A$
where $Z$ is an $ntimes n$ matrix, $B$ is an $m times m$ matrix, and $A$ is a $m times n$ matrix, $n<m$.
Furthermore, the columns of $A$ are orthogonal, and $B$ is a positive definite matrix and $Z$ is symmetric tridiagonal.
This bears similarities to two similar $mtimes m$ matrices, $X$ and $Y$, requiring Y = $P^{-1}XP$ for some invertible matrix $P$.
Obviously in my case, $B$ and $Z$ don't have the same dimensions, and as $A$ is a truncated orthogonal matrix, it only has the property $A^TA=I$ (and not the other way round).
However, can any properties of similar matrices follow from this, such as the characteristic polynomial, determinant or eigenvalues?
linear-algebra abstract-algebra matrices
linear-algebra abstract-algebra matrices
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
asked Nov 30 '18 at 17:55
Thomas SmithThomas Smith
685
685
add a comment |
add a comment |
1 Answer
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$begingroup$
Extend $A$ to an orthogonal matrix $Q=pmatrix{A&ast}$. Then $Z$ is the leading principal $ntimes n$ submatrix of $Q^TBQ$. Therefore $Z$ is positive definite and by Cauchy interlacing inequality, the $i$-th smallest eigenvalue of $Z$ is bounded below by the $i$-th smallest eigenvalue of $B$.
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
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add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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active
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votes
$begingroup$
Extend $A$ to an orthogonal matrix $Q=pmatrix{A&ast}$. Then $Z$ is the leading principal $ntimes n$ submatrix of $Q^TBQ$. Therefore $Z$ is positive definite and by Cauchy interlacing inequality, the $i$-th smallest eigenvalue of $Z$ is bounded below by the $i$-th smallest eigenvalue of $B$.
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
$endgroup$
add a comment |
$begingroup$
Extend $A$ to an orthogonal matrix $Q=pmatrix{A&ast}$. Then $Z$ is the leading principal $ntimes n$ submatrix of $Q^TBQ$. Therefore $Z$ is positive definite and by Cauchy interlacing inequality, the $i$-th smallest eigenvalue of $Z$ is bounded below by the $i$-th smallest eigenvalue of $B$.
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
$endgroup$
add a comment |
$begingroup$
Extend $A$ to an orthogonal matrix $Q=pmatrix{A&ast}$. Then $Z$ is the leading principal $ntimes n$ submatrix of $Q^TBQ$. Therefore $Z$ is positive definite and by Cauchy interlacing inequality, the $i$-th smallest eigenvalue of $Z$ is bounded below by the $i$-th smallest eigenvalue of $B$.
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
$endgroup$
Extend $A$ to an orthogonal matrix $Q=pmatrix{A&ast}$. Then $Z$ is the leading principal $ntimes n$ submatrix of $Q^TBQ$. Therefore $Z$ is positive definite and by Cauchy interlacing inequality, the $i$-th smallest eigenvalue of $Z$ is bounded below by the $i$-th smallest eigenvalue of $B$.
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
answered Nov 30 '18 at 18:19
user1551user1551
72k566126
72k566126
add a comment |
add a comment |
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