Mixing
$clambda$ is an eigenvalue of $cA$?
$begingroup$
"Let $A$ a square matrix have an eigenvalue $lambda$ with corresponding eigenvector $v$. Let $cin F$. Then $clambda$ is an eigenvalue of $cA$."
This is true right? since the eigenvalues of $A$ form the eigenvectors in $A$, multiplying $A$ by a scalar $c$ is still equivalent to $A$, so their eigenvalues must be proportional too?
linear-algebra
edited Jan 21 at 1:59
El borito
666216
asked Jan 21 at 1:50
ximxim
516
$endgroup$
add a comment |
$begingroup$
"Let $A$ a square matrix have an eigenvalue $lambda$ with corresponding eigenvector $v$. Let $cin F$. Then $clambda$ is an eigenvalue of $cA$."
This is true right? since the eigenvalues of $A$ form the eigenvectors in $A$, multiplying $A$ by a scalar $c$ is still equivalent to $A$, so their eigenvalues must be proportional too?
linear-algebra
edited Jan 21 at 1:59
El borito
666216
asked Jan 21 at 1:50
ximxim
516
$endgroup$
1
$begingroup$
More concretely, simply show that there is an eigenvector of $cA$ with eigenvalue $clambda$...
$endgroup$
– angryavian
Jan 21 at 1:51
add a comment |
$begingroup$
"Let $A$ a square matrix have an eigenvalue $lambda$ with corresponding eigenvector $v$. Let $cin F$. Then $clambda$ is an eigenvalue of $cA$."
This is true right? since the eigenvalues of $A$ form the eigenvectors in $A$, multiplying $A$ by a scalar $c$ is still equivalent to $A$, so their eigenvalues must be proportional too?
linear-algebra
edited Jan 21 at 1:59
El borito
666216
asked Jan 21 at 1:50
ximxim
516
$endgroup$
"Let $A$ a square matrix have an eigenvalue $lambda$ with corresponding eigenvector $v$. Let $cin F$. Then $clambda$ is an eigenvalue of $cA$."
This is true right? since the eigenvalues of $A$ form the eigenvectors in $A$, multiplying $A$ by a scalar $c$ is still equivalent to $A$, so their eigenvalues must be proportional too?
linear-algebra
linear-algebra
edited Jan 21 at 1:59
El borito
666216
asked Jan 21 at 1:50
ximxim
516
edited Jan 21 at 1:59
El borito
666216
asked Jan 21 at 1:50
ximxim
516
edited Jan 21 at 1:59
El borito
666216
edited Jan 21 at 1:59
El borito
666216
edited Jan 21 at 1:59
El borito
666216
666216
asked Jan 21 at 1:50
ximxim
516
asked Jan 21 at 1:50
ximxim
516
asked Jan 21 at 1:50
ximxim
516
516
1
$begingroup$
More concretely, simply show that there is an eigenvector of $cA$ with eigenvalue $clambda$...
$endgroup$
– angryavian
Jan 21 at 1:51
add a comment |
1
$begingroup$
More concretely, simply show that there is an eigenvector of $cA$ with eigenvalue $clambda$...
$endgroup$
– angryavian
Jan 21 at 1:51
1
1
$begingroup$
More concretely, simply show that there is an eigenvector of $cA$ with eigenvalue $clambda$...
$endgroup$
– angryavian
Jan 21 at 1:51
$begingroup$
More concretely, simply show that there is an eigenvector of $cA$ with eigenvalue $clambda$...
$endgroup$
– angryavian
Jan 21 at 1:51
add a comment |
1 Answer
1
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$begingroup$
Yes, this is easy to check: we have $$(cA)v=c(Av)=c(lambda v)=(clambda)v.$$ The first equality is by definition, the second by the assumption that $v$ is an eigenvector of $A$ with eigenvalue $lambda$, and the third equality is just the basic associativity property of scalar multiplication.
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
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$begingroup$
Yes, this is easy to check: we have $$(cA)v=c(Av)=c(lambda v)=(clambda)v.$$ The first equality is by definition, the second by the assumption that $v$ is an eigenvector of $A$ with eigenvalue $lambda$, and the third equality is just the basic associativity property of scalar multiplication.
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
Yes, this is easy to check: we have $$(cA)v=c(Av)=c(lambda v)=(clambda)v.$$ The first equality is by definition, the second by the assumption that $v$ is an eigenvector of $A$ with eigenvalue $lambda$, and the third equality is just the basic associativity property of scalar multiplication.
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
Yes, this is easy to check: we have $$(cA)v=c(Av)=c(lambda v)=(clambda)v.$$ The first equality is by definition, the second by the assumption that $v$ is an eigenvector of $A$ with eigenvalue $lambda$, and the third equality is just the basic associativity property of scalar multiplication.
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
$endgroup$
Yes, this is easy to check: we have $$(cA)v=c(Av)=c(lambda v)=(clambda)v.$$ The first equality is by definition, the second by the assumption that $v$ is an eigenvector of $A$ with eigenvalue $lambda$, and the third equality is just the basic associativity property of scalar multiplication.
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
answered Jan 21 at 1:53
Noah SchweberNoah Schweber
126k10151290
126k10151290
add a comment |
add a comment |
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$begingroup$
More concretely, simply show that there is an eigenvector of $cA$ with eigenvalue $clambda$...
$endgroup$
– angryavian
Jan 21 at 1:51