Sum of symmetric, positive semidefinite matricesDoes this conic combination generate all $ntimes n$ real symmetric positive-semidefinite matrices?A is symmetric iff A=P-Q, where P,Q are positive definite matricesCriterion for positive semidefinite matricesAnother property of symmetric positive semidefinite matricesSum of rank-$k$ positive semidefinite matrices is at least of rank $k$Simultaneous Diagonalization of Symmetric Positive Semidefinite matricesA question about Hermitian and positive semidefinite matricesIs this matrix product positive semidefinite?Sum of rank 1 positive semidefinite and negative semidefinite matricesPositive semidefinite matrix using Schur Complement
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Sum of symmetric, positive semidefinite matrices
Does this conic combination generate all $ntimes n$ real symmetric positive-semidefinite matrices?A is symmetric iff A=P-Q, where P,Q are positive definite matricesCriterion for positive semidefinite matricesAnother property of symmetric positive semidefinite matricesSum of rank-$k$ positive semidefinite matrices is at least of rank $k$Simultaneous Diagonalization of Symmetric Positive Semidefinite matricesA question about Hermitian and positive semidefinite matricesIs this matrix product positive semidefinite?Sum of rank 1 positive semidefinite and negative semidefinite matricesPositive semidefinite matrix using Schur Complement
$begingroup$
Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.
I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?
linear-algebra matrices symmetric-matrices positive-semidefinite
edited yesterday
Rodrigo de Azevedo
13k41960
asked yesterday
mich95mich95
6,97211126
$endgroup$
add a comment |
$begingroup$
Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.
I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?
linear-algebra matrices symmetric-matrices positive-semidefinite
edited yesterday
Rodrigo de Azevedo
13k41960
asked yesterday
mich95mich95
6,97211126
$endgroup$
$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
$begingroup$
Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.
I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?
linear-algebra matrices symmetric-matrices positive-semidefinite
edited yesterday
Rodrigo de Azevedo
13k41960
asked yesterday
mich95mich95
6,97211126
$endgroup$
Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.
I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?
linear-algebra matrices symmetric-matrices positive-semidefinite
linear-algebra matrices symmetric-matrices positive-semidefinite
edited yesterday
Rodrigo de Azevedo
13k41960
asked yesterday
mich95mich95
6,97211126
edited yesterday
Rodrigo de Azevedo
13k41960
asked yesterday
mich95mich95
6,97211126
edited yesterday
Rodrigo de Azevedo
13k41960
edited yesterday
Rodrigo de Azevedo
13k41960
edited yesterday
Rodrigo de Azevedo
13k41960
13k41960
asked yesterday
mich95mich95
6,97211126
asked yesterday
mich95mich95
6,97211126
asked yesterday
mich95mich95
6,97211126
6,97211126
$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday
$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
$endgroup$
add a comment |
$begingroup$
Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
$$
0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
$$
conclude that $x in ker(A) cap ker(B)$.
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
$endgroup$
add a comment |
$begingroup$
$newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
$endgroup$
add a comment |
$begingroup$
$newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
$endgroup$
$newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
edited yesterday
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
answered yesterday
Minus One-TwelfthMinus One-Twelfth
2,07219
2,07219
add a comment |
add a comment |
$begingroup$
Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
$$
0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
$$
conclude that $x in ker(A) cap ker(B)$.
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
add a comment |
$begingroup$
Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
$$
0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
$$
conclude that $x in ker(A) cap ker(B)$.
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
add a comment |
$begingroup$
Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
$$
0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
$$
conclude that $x in ker(A) cap ker(B)$.
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
$endgroup$
Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
$$
0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
$$
conclude that $x in ker(A) cap ker(B)$.
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
answered yesterday
OmnomnomnomOmnomnomnom
128k791186
128k791186
add a comment |
add a comment |
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$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday