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Problem

The matrix $A \in \mathbb{R}^{4\times4}$ has eigenvectors $\bf{u_1,u_2,u_3,u_4}$ where $\bf{u_1}=\begin{pmatrix}1\\1\\0\\1\end{pmatrix}$, $\bf{u_2}=\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$, $\bf{u_3}=\begin{pmatrix}1\\-1\\0\\1\end{pmatrix}$ satisfy $A\bf{u_1=5u_1}$, $A\bf{u_2=9u_2}$ and $A\bf{u_3=20u_3}$.
Calculate $A\bf{w}$ where $\bf{w}=\begin{pmatrix}13\\7\\12\\13\end{pmatrix}$

At first I thought I should use $A=PDP^{-1}$ , where $P$ is eigenvector matrix and $D$ is eigenvalue matrix. If I'm not mistaken, from the question the eigenvalues are $\lambda_1=5$, $\lambda_2=9$ and $\lambda_3=20$ right? But I don't have the $\bf{u_4}$ and $\lambda_4$.

Do I need to know all the eigenvalues and eigenvectors? Do I need to find these values to calculate $A$ or is there another method?

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  • $\begingroup$ note that $u3= <1,-1,0,1>$ $\endgroup$
    – user
    Jan 17, 2018 at 22:24
  • $\begingroup$ math.stackexchange.com/questions/2604750 $\endgroup$
    – Angina Seng
    Jan 17, 2018 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, 2018 at 22:28

2 Answers 2

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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: $b=12, a+c=1, a-c=-5$

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  • $\begingroup$ After i found a,b and c how exactly can i calculate Aw? Am i missing something? $\endgroup$
    – user522473
    Jan 17, 2018 at 22:34
  • $\begingroup$ @Chloe by linearity $$Aw=A(a\cdot u_1+b\cdot u_2+c\cdot u_3)=a\cdot A u_1+b\cdot Au_2+c\cdot Au_3$$ $\endgroup$
    – user
    Jan 17, 2018 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, 2018 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 $u_1$, $u_2$, $u_3$. $\endgroup$
    – user
    Jan 17, 2018 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 $Au_i$. $\endgroup$
    – user
    Jan 17, 2018 at 22:46
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Here $w$ is not a linear combination of $u_1, u_2, u_3$ so with this imformation it is not possible to calculate $Aw$.

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  • $\begingroup$ In fact, it is a linear combination of those vectors. $\endgroup$
    – amd
    Jan 17, 2018 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, 2018 at 0:14
  • $\begingroup$ I’m sure you can, but as things stand now, your answer is incorrect. $\endgroup$
    – amd
    Jan 18, 2018 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, 2018 at 0:45
  • $\begingroup$ Anyway you are right. $\endgroup$
    – GhD
    Jan 18, 2018 at 0:46

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