# Matrix $A\in \mathbb{R}^{4\times4}$ has eigenvectors $\bf{u_1,u_2,u_3,u_4}$ satisfying $\bf{Au_1=5u_1,Au_2=9u_2}$ & $\bf{Au_3=20u_3}$. Find $A\bf{w}$.

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?

• note that $u3= <1,-1,0,1>$
– user
Jan 17, 2018 at 22:24
• math.stackexchange.com/questions/2604750 Jan 17, 2018 at 22:25
• 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 $\frac {c_1}{\lambda_1}, \frac {c_2}{\lambda_2},\frac {c_3}{\lambda_3}$ Jan 17, 2018 at 22:28

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

• After i found a,b and c how exactly can i calculate Aw? Am i missing something?
– user522473
Jan 17, 2018 at 22:34
• @Chloe by linearity $$Aw=A(a\cdot u_1+b\cdot u_2+c\cdot u_3)=a\cdot A u_1+b\cdot Au_2+c\cdot Au_3$$
– user
Jan 17, 2018 at 22:35
• Thank you! Just one thing how is it that w=a⋅u1+b⋅u2+c⋅u3 , is this a rule im forgetting? thanks again
– user522473
Jan 17, 2018 at 22:40
• @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 $u_1$, $u_2$, $u_3$.
– user
Jan 17, 2018 at 22:44
• @Chloe Note also that it is not important that $u_1$, $u_2$, $u_3$ are eigenvectors, what we need is to know $Au_i$.
– user
Jan 17, 2018 at 22:46

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

• In fact, it is a linear combination of those vectors.
– amd
Jan 17, 2018 at 22:46
• @amd Now yes, he/she changed $u_3$. Before, the first element of $u_3$ was $-1$. I can calculate such easy thing!
– GhD
Jan 18, 2018 at 0:14
• I’m sure you can, but as things stand now, your answer is incorrect.
– amd
Jan 18, 2018 at 0:38
• @amd Yes, but if you see the editing history you will understand the first element of $u_3$ was $-1$ :)
– GhD
Jan 18, 2018 at 0:45
• Anyway you are right.
– GhD
Jan 18, 2018 at 0:46