Suppose a square matrix $A$ has spectral radius $rho(A) < 1$. Fixing the last row and scaling other entries by $r in (0,1)$, will $rho(A)<1$? The 2019 Stack Overflow Developer Survey Results Are InDifferent form of determinant, does it make mine wrong?Spectral radius of the product of a right stochastic matrix and a block diagonal matrixSpectral radius of the product of a right stochastic matrix and hermitian matrixLower bound spectral radius of matrixUpper bound of spectral radius of the sum of two matrices, one with spectral radius no larger than 1, and the other has small eigenvaluesSpetral properties of differential of Lyapunov matrix equationSolving Vandermonde equation systemDoes a matrix of minimum norm in an affine subspace of $M_n(mathbb R)$ have minimum spectral radius?Are the sets $X: max_i textRelambda_i (B+AX) < 0$ and $X: rho(B+AX) < 1$ homeomorphic?Is the function $A mapsto sumlimits_j=0^infty langle A^j v, A^j v rangle$ differentiable everywhere?
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Suppose a square matrix $A$ has spectral radius $rho(A)
The 2019 Stack Overflow Developer Survey Results Are InDifferent form of determinant, does it make mine wrong?Spectral radius of the product of a right stochastic matrix and a block diagonal matrixSpectral radius of the product of a right stochastic matrix and hermitian matrixLower bound spectral radius of matrixUpper bound of spectral radius of the sum of two matrices, one with spectral radius no larger than 1, and the other has small eigenvaluesSpetral properties of differential of Lyapunov matrix equationSolving Vandermonde equation systemDoes a matrix of minimum norm in an affine subspace of $M_n(mathbb R)$ have minimum spectral radius?Are the sets $X: max_i textRelambda_i (B+AX) < 0$ and $X: rho(B+AX) < 1$ homeomorphic?Is the function $A mapsto sumlimits_j=0^infty langle A^j v, A^j v rangle$ differentiable everywhere?
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Suppose $A in M_n(mathbb R)$ has spectral radius small than $1$, i.e., $rho(A) < 1$. Denote $A = pmatrixa_1^T \ vdots \ a_n^T$, where $a_j^T$ denotes the $j^th$ row of $A$.Putting $B=pmatrixa_1^T \ vdots \ a_n-1^T$, i.e., the first $n-1$ rows of $A$, if $r in (0, 1)$ what should the spectral radius of $(A)_r = pmatrixrB \ a_n^T$ be? That is, $(A)_r$ is the matrix we fix the last row of $A$ but scale other rows by a factor $r$. I feel that $rho( (A)_r)$ should be smaller than $1$ too but could not prove it or give a counterexample.
linear-algebra spectral-radius matrix-analysis
$endgroup$
add a comment |
$begingroup$
Suppose $A in M_n(mathbb R)$ has spectral radius small than $1$, i.e., $rho(A) < 1$. Denote $A = pmatrixa_1^T \ vdots \ a_n^T$, where $a_j^T$ denotes the $j^th$ row of $A$.Putting $B=pmatrixa_1^T \ vdots \ a_n-1^T$, i.e., the first $n-1$ rows of $A$, if $r in (0, 1)$ what should the spectral radius of $(A)_r = pmatrixrB \ a_n^T$ be? That is, $(A)_r$ is the matrix we fix the last row of $A$ but scale other rows by a factor $r$. I feel that $rho( (A)_r)$ should be smaller than $1$ too but could not prove it or give a counterexample.
linear-algebra spectral-radius matrix-analysis
$endgroup$
add a comment |
$begingroup$
Suppose $A in M_n(mathbb R)$ has spectral radius small than $1$, i.e., $rho(A) < 1$. Denote $A = pmatrixa_1^T \ vdots \ a_n^T$, where $a_j^T$ denotes the $j^th$ row of $A$.Putting $B=pmatrixa_1^T \ vdots \ a_n-1^T$, i.e., the first $n-1$ rows of $A$, if $r in (0, 1)$ what should the spectral radius of $(A)_r = pmatrixrB \ a_n^T$ be? That is, $(A)_r$ is the matrix we fix the last row of $A$ but scale other rows by a factor $r$. I feel that $rho( (A)_r)$ should be smaller than $1$ too but could not prove it or give a counterexample.
linear-algebra spectral-radius matrix-analysis
$endgroup$
Suppose $A in M_n(mathbb R)$ has spectral radius small than $1$, i.e., $rho(A) < 1$. Denote $A = pmatrixa_1^T \ vdots \ a_n^T$, where $a_j^T$ denotes the $j^th$ row of $A$.Putting $B=pmatrixa_1^T \ vdots \ a_n-1^T$, i.e., the first $n-1$ rows of $A$, if $r in (0, 1)$ what should the spectral radius of $(A)_r = pmatrixrB \ a_n^T$ be? That is, $(A)_r$ is the matrix we fix the last row of $A$ but scale other rows by a factor $r$. I feel that $rho( (A)_r)$ should be smaller than $1$ too but could not prove it or give a counterexample.
linear-algebra spectral-radius matrix-analysis
linear-algebra spectral-radius matrix-analysis
asked Mar 24 at 5:21
user1101010user1101010
9011830
9011830
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1 Answer
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The answer is no. Let $A$ be any nilpotent matrix such that $|a_nn|>1$, such as $A=alphapmatrix-1&-1\ 1&1$ for any $alpha>1$. Then $rho(A)=0$ but $rho(A_r)to|a_nn|>1$ when $rto0$.
However, the answer is yes if $A$ is (entrywise) nonnegative. In this case, as $0le A_r^kle A^k$ for all positive integer $k$, by using Gelfand's formula with Frobenius norm, we have
$$
rho(A_r)=lim_ktoinfty|A_r^k|^1/klelim_ktoinfty|A^k|^1/k=rho(A)<1.
$$
$endgroup$
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answer is no. Let $A$ be any nilpotent matrix such that $|a_nn|>1$, such as $A=alphapmatrix-1&-1\ 1&1$ for any $alpha>1$. Then $rho(A)=0$ but $rho(A_r)to|a_nn|>1$ when $rto0$.
However, the answer is yes if $A$ is (entrywise) nonnegative. In this case, as $0le A_r^kle A^k$ for all positive integer $k$, by using Gelfand's formula with Frobenius norm, we have
$$
rho(A_r)=lim_ktoinfty|A_r^k|^1/klelim_ktoinfty|A^k|^1/k=rho(A)<1.
$$
$endgroup$
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
add a comment |
$begingroup$
The answer is no. Let $A$ be any nilpotent matrix such that $|a_nn|>1$, such as $A=alphapmatrix-1&-1\ 1&1$ for any $alpha>1$. Then $rho(A)=0$ but $rho(A_r)to|a_nn|>1$ when $rto0$.
However, the answer is yes if $A$ is (entrywise) nonnegative. In this case, as $0le A_r^kle A^k$ for all positive integer $k$, by using Gelfand's formula with Frobenius norm, we have
$$
rho(A_r)=lim_ktoinfty|A_r^k|^1/klelim_ktoinfty|A^k|^1/k=rho(A)<1.
$$
$endgroup$
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
add a comment |
$begingroup$
The answer is no. Let $A$ be any nilpotent matrix such that $|a_nn|>1$, such as $A=alphapmatrix-1&-1\ 1&1$ for any $alpha>1$. Then $rho(A)=0$ but $rho(A_r)to|a_nn|>1$ when $rto0$.
However, the answer is yes if $A$ is (entrywise) nonnegative. In this case, as $0le A_r^kle A^k$ for all positive integer $k$, by using Gelfand's formula with Frobenius norm, we have
$$
rho(A_r)=lim_ktoinfty|A_r^k|^1/klelim_ktoinfty|A^k|^1/k=rho(A)<1.
$$
$endgroup$
The answer is no. Let $A$ be any nilpotent matrix such that $|a_nn|>1$, such as $A=alphapmatrix-1&-1\ 1&1$ for any $alpha>1$. Then $rho(A)=0$ but $rho(A_r)to|a_nn|>1$ when $rto0$.
However, the answer is yes if $A$ is (entrywise) nonnegative. In this case, as $0le A_r^kle A^k$ for all positive integer $k$, by using Gelfand's formula with Frobenius norm, we have
$$
rho(A_r)=lim_ktoinfty|A_r^k|^1/klelim_ktoinfty|A^k|^1/k=rho(A)<1.
$$
edited Mar 24 at 6:15
answered Mar 24 at 6:01
user1551user1551
74.3k566129
74.3k566129
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
add a comment |
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
$begingroup$
Thanks for your example. Could you see some sufficient conditions on $A$ for such claim ($rho(A_r) < 1$) to hold?
$endgroup$
– user1101010
Mar 24 at 6:07
add a comment |
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