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 ??










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  • 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















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 ??










share|cite|improve this question


















  • 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













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 ??










share|cite|improve this question













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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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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















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