Let $A$ and $B$ be real matrix such that $A+iB$ is non singular show that there exist $t in mathbbR$ such that $A+tB$ is non singular

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Let $A$ and $B$ be real matrices, with $A+iB$ non-singular. I need to show that there exist a real number $t$ such that $ A+tB $ is non-singular.



I don't have any idea how I can approach this question... could I please get a hint?






linear-algebra







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  • What did you try?
    – José Carlos Santos
    Aug 3 at 16:55






  • 5




    I don't believe that you don't have any idea. You know more than you think. The question asks about non-singular matrices. What do you know about those? How would you most easily tell whether a matrix is singular or not? Have you tried to apply that here?
    – Arthur
    Aug 3 at 16:57







  • 8




    $p(t)=det(A+tB)$ is a polynomial with real coefficients. It is a non-zero polynomial, because $p(i)neq0$. Therefore, it cannot be zero for all values of $t$.
    – user580373
    Aug 3 at 17:04










  • @spiralstotheleft / That is, it cannot be $0$ for all real $t$.
    – DanielWainfleet
    Aug 4 at 1:34










  • I wonder if there is a geometric proof of this?
    – copper.hat
    Aug 6 at 2:20














up vote
5
down vote

favorite












Let $A$ and $B$ be real matrices, with $A+iB$ non-singular. I need to show that there exist a real number $t$ such that $ A+tB $ is non-singular.



I don't have any idea how I can approach this question... could I please get a hint?






linear-algebra







share|cite|improve this question























  • What did you try?
    – José Carlos Santos
    Aug 3 at 16:55






  • 5




    I don't believe that you don't have any idea. You know more than you think. The question asks about non-singular matrices. What do you know about those? How would you most easily tell whether a matrix is singular or not? Have you tried to apply that here?
    – Arthur
    Aug 3 at 16:57







  • 8




    $p(t)=det(A+tB)$ is a polynomial with real coefficients. It is a non-zero polynomial, because $p(i)neq0$. Therefore, it cannot be zero for all values of $t$.
    – user580373
    Aug 3 at 17:04










  • @spiralstotheleft / That is, it cannot be $0$ for all real $t$.
    – DanielWainfleet
    Aug 4 at 1:34










  • I wonder if there is a geometric proof of this?
    – copper.hat
    Aug 6 at 2:20












up vote
5
down vote

favorite









up vote
5
down vote

favorite











Let $A$ and $B$ be real matrices, with $A+iB$ non-singular. I need to show that there exist a real number $t$ such that $ A+tB $ is non-singular.



I don't have any idea how I can approach this question... could I please get a hint?






linear-algebra







share|cite|improve this question















Let $A$ and $B$ be real matrices, with $A+iB$ non-singular. I need to show that there exist a real number $t$ such that $ A+tB $ is non-singular.



I don't have any idea how I can approach this question... could I please get a hint?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 1 at 8:55









Jneven

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633320










asked Aug 3 at 16:53









Renu

311




311











  • What did you try?
    – José Carlos Santos
    Aug 3 at 16:55






  • 5




    I don't believe that you don't have any idea. You know more than you think. The question asks about non-singular matrices. What do you know about those? How would you most easily tell whether a matrix is singular or not? Have you tried to apply that here?
    – Arthur
    Aug 3 at 16:57







  • 8




    $p(t)=det(A+tB)$ is a polynomial with real coefficients. It is a non-zero polynomial, because $p(i)neq0$. Therefore, it cannot be zero for all values of $t$.
    – user580373
    Aug 3 at 17:04










  • @spiralstotheleft / That is, it cannot be $0$ for all real $t$.
    – DanielWainfleet
    Aug 4 at 1:34










  • I wonder if there is a geometric proof of this?
    – copper.hat
    Aug 6 at 2:20
















  • What did you try?
    – José Carlos Santos
    Aug 3 at 16:55






  • 5




    I don't believe that you don't have any idea. You know more than you think. The question asks about non-singular matrices. What do you know about those? How would you most easily tell whether a matrix is singular or not? Have you tried to apply that here?
    – Arthur
    Aug 3 at 16:57







  • 8




    $p(t)=det(A+tB)$ is a polynomial with real coefficients. It is a non-zero polynomial, because $p(i)neq0$. Therefore, it cannot be zero for all values of $t$.
    – user580373
    Aug 3 at 17:04










  • @spiralstotheleft / That is, it cannot be $0$ for all real $t$.
    – DanielWainfleet
    Aug 4 at 1:34










  • I wonder if there is a geometric proof of this?
    – copper.hat
    Aug 6 at 2:20















What did you try?
– José Carlos Santos
Aug 3 at 16:55




What did you try?
– José Carlos Santos
Aug 3 at 16:55




5




5




I don't believe that you don't have any idea. You know more than you think. The question asks about non-singular matrices. What do you know about those? How would you most easily tell whether a matrix is singular or not? Have you tried to apply that here?
– Arthur
Aug 3 at 16:57





I don't believe that you don't have any idea. You know more than you think. The question asks about non-singular matrices. What do you know about those? How would you most easily tell whether a matrix is singular or not? Have you tried to apply that here?
– Arthur
Aug 3 at 16:57





8




8




$p(t)=det(A+tB)$ is a polynomial with real coefficients. It is a non-zero polynomial, because $p(i)neq0$. Therefore, it cannot be zero for all values of $t$.
– user580373
Aug 3 at 17:04




$p(t)=det(A+tB)$ is a polynomial with real coefficients. It is a non-zero polynomial, because $p(i)neq0$. Therefore, it cannot be zero for all values of $t$.
– user580373
Aug 3 at 17:04












@spiralstotheleft / That is, it cannot be $0$ for all real $t$.
– DanielWainfleet
Aug 4 at 1:34




@spiralstotheleft / That is, it cannot be $0$ for all real $t$.
– DanielWainfleet
Aug 4 at 1:34












I wonder if there is a geometric proof of this?
– copper.hat
Aug 6 at 2:20




I wonder if there is a geometric proof of this?
– copper.hat
Aug 6 at 2:20















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