0
$\begingroup$

QUESTIONThe $4\times4$ matrix $A$ has eigenvectors $u_1,u_2,u_3, $ and $u_4$. I've been give $3$ eigenvectors $u_1= <1,1,0,1> u_2= <1,1,1,1> u_3= <1,-1,0,1>$ and have been told that they satisfy $Au_1=5u_1,Au_2=9u_2,$ and $Au_3=20u_3$. I've been asked to calculate $Aw$ where $w= <13,7,12,13>$

At first I thought I should use $A=PDP^{-1}$ , where $P$= eigenvector matrix and D= eigenvalue matrix. If I'm not mistaken, from the question the eigenvalues are $λ_1=5$, $λ_2=9$ and $λ_3=20$ right? But I don't have the $u_4$ and $λ_4$. Do I need to find these values to calculate $A$ or is there another method?

Thank you for your help

$\endgroup$
  • $\begingroup$ note that $u3= <1,-1,0,1>$ $\endgroup$ – gimusi Jan 17 '18 at 22:24
  • $\begingroup$ math.stackexchange.com/questions/2604750 $\endgroup$ – Lord Shark the Unknown Jan 17 '18 at 22:25
  • $\begingroup$ In this case you do not need to find $u_4.$ Find $\{c_1,c_2,c_3\}$ such that $c_1u_1 + c_2 u_2 + c_3 u_3 = w$ then $\frac {c_1}{\lambda_1}, \frac {c_2}{\lambda_2},\frac {c_3}{\lambda_3}$ $\endgroup$ – Doug M Jan 17 '18 at 22:28
1
$\begingroup$

HINT

We can calculate $Aw$ only if we can find $a,b,c$ such that

$$w=a\cdot u_1+b\cdot u_2+c\cdot u_3$$

thus you can easily find that: $b=12, a+c=1, a-c=-5$

$\endgroup$
  • $\begingroup$ After i found a,b and c how exactly can i calculate Aw? Am i missing something? $\endgroup$ – user522473 Jan 17 '18 at 22:34
  • $\begingroup$ @Chloe by linearity $$Aw=A(a\cdot u_1+b\cdot u_2+c\cdot u_3)=a\cdot A u_1+b\cdot Au_2+c\cdot Au_3$$ $\endgroup$ – gimusi Jan 17 '18 at 22:35
  • $\begingroup$ Thank you! Just one thing how is it that w=a⋅u1+b⋅u2+c⋅u3 , is this a rule im forgetting? thanks again $\endgroup$ – user522473 Jan 17 '18 at 22:40
  • $\begingroup$ @Chloe We don't have sufficient information to know completely the matrix A, we only know the result for $Au_1$, $Au_2$ and $Au_3$ thus we cal calculate Aw with if and only if w is a linear combination of $u_1$, $u_2$, $u_3$. $\endgroup$ – gimusi Jan 17 '18 at 22:44
  • $\begingroup$ @Chloe Note also that it is not important that $u_1$, $u_2$, $u_3$ are eigenvectors, what we need is to know $Au_i$. $\endgroup$ – gimusi Jan 17 '18 at 22:46
0
$\begingroup$

Here $w$ is not a linear combination of $u_1, u_2, u_3$ so with this imformation it is not possible to calculate $Aw$.

$\endgroup$
  • $\begingroup$ In fact, it is a linear combination of those vectors. $\endgroup$ – amd Jan 17 '18 at 22:46
  • $\begingroup$ @amd Now yes, he/she changed $u_3$. Before, the first element of $u_3$ was $-1$. I can calculate such easy thing! $\endgroup$ – GhD Jan 18 '18 at 0:14
  • $\begingroup$ I’m sure you can, but as things stand now, your answer is incorrect. $\endgroup$ – amd Jan 18 '18 at 0:38
  • $\begingroup$ @amd Yes, but if you see the editing history you will understand the first element of $u_3$ was $-1$ :) $\endgroup$ – GhD Jan 18 '18 at 0:45
  • $\begingroup$ Anyway you are right. $\endgroup$ – GhD Jan 18 '18 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy