If $lambda$ is an eignevalue of $B$,then choose the correct option
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let $A$ be any $ntimes n$ non singular complex matrix and let $B = (bar A)^tA,$ where $(bar A)^t$ is the conjugate transpose of $A$. If $lambda$ is an eignevalue of $B$,then
choose the correct option
$a)$$lambda$ is real and $lambda < 0$
$b)$$lambda$ is real and $lambda le 0$
$c)$$lambda$ is real and $lambda > 0$
$d)$$lambda$ is real and $lambda ge 0$
I thinks option $d)$ will correct
Is it True ??
linear-algebra
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0
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let $A$ be any $ntimes n$ non singular complex matrix and let $B = (bar A)^tA,$ where $(bar A)^t$ is the conjugate transpose of $A$. If $lambda$ is an eignevalue of $B$,then
choose the correct option
$a)$$lambda$ is real and $lambda < 0$
$b)$$lambda$ is real and $lambda le 0$
$c)$$lambda$ is real and $lambda > 0$
$d)$$lambda$ is real and $lambda ge 0$
I thinks option $d)$ will correct
Is it True ??
linear-algebra
2
For this kind of exercises, you can take a rather good hint from assuming stuff on $;A;$ . For example, what if this matrix , and thus also $;B;$ , is singular? Then it could perfectly well be that $;lambda=0;$ ...Can you see how the number of options gets smaller?
– DonAntonio
Nov 17 at 21:45
@DonAntonio thanks u...i missed that
– Messi fifa
Nov 17 at 21:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
let $A$ be any $ntimes n$ non singular complex matrix and let $B = (bar A)^tA,$ where $(bar A)^t$ is the conjugate transpose of $A$. If $lambda$ is an eignevalue of $B$,then
choose the correct option
$a)$$lambda$ is real and $lambda < 0$
$b)$$lambda$ is real and $lambda le 0$
$c)$$lambda$ is real and $lambda > 0$
$d)$$lambda$ is real and $lambda ge 0$
I thinks option $d)$ will correct
Is it True ??
linear-algebra
let $A$ be any $ntimes n$ non singular complex matrix and let $B = (bar A)^tA,$ where $(bar A)^t$ is the conjugate transpose of $A$. If $lambda$ is an eignevalue of $B$,then
choose the correct option
$a)$$lambda$ is real and $lambda < 0$
$b)$$lambda$ is real and $lambda le 0$
$c)$$lambda$ is real and $lambda > 0$
$d)$$lambda$ is real and $lambda ge 0$
I thinks option $d)$ will correct
Is it True ??
linear-algebra
linear-algebra
asked Nov 17 at 21:41
Messi fifa
50111
50111
2
For this kind of exercises, you can take a rather good hint from assuming stuff on $;A;$ . For example, what if this matrix , and thus also $;B;$ , is singular? Then it could perfectly well be that $;lambda=0;$ ...Can you see how the number of options gets smaller?
– DonAntonio
Nov 17 at 21:45
@DonAntonio thanks u...i missed that
– Messi fifa
Nov 17 at 21:46
add a comment |
2
For this kind of exercises, you can take a rather good hint from assuming stuff on $;A;$ . For example, what if this matrix , and thus also $;B;$ , is singular? Then it could perfectly well be that $;lambda=0;$ ...Can you see how the number of options gets smaller?
– DonAntonio
Nov 17 at 21:45
@DonAntonio thanks u...i missed that
– Messi fifa
Nov 17 at 21:46
2
2
For this kind of exercises, you can take a rather good hint from assuming stuff on $;A;$ . For example, what if this matrix , and thus also $;B;$ , is singular? Then it could perfectly well be that $;lambda=0;$ ...Can you see how the number of options gets smaller?
– DonAntonio
Nov 17 at 21:45
For this kind of exercises, you can take a rather good hint from assuming stuff on $;A;$ . For example, what if this matrix , and thus also $;B;$ , is singular? Then it could perfectly well be that $;lambda=0;$ ...Can you see how the number of options gets smaller?
– DonAntonio
Nov 17 at 21:45
@DonAntonio thanks u...i missed that
– Messi fifa
Nov 17 at 21:46
@DonAntonio thanks u...i missed that
– Messi fifa
Nov 17 at 21:46
add a comment |
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For this kind of exercises, you can take a rather good hint from assuming stuff on $;A;$ . For example, what if this matrix , and thus also $;B;$ , is singular? Then it could perfectly well be that $;lambda=0;$ ...Can you see how the number of options gets smaller?
– DonAntonio
Nov 17 at 21:45
@DonAntonio thanks u...i missed that
– Messi fifa
Nov 17 at 21:46