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I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.

I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector. Any help is appreciated, thanks!

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    $\begingroup$ Different eigenspaces of a symmetric matrix are orthogonal to each other. $\endgroup$
    – Berci
    Dec 7 '18 at 21:54
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    $\begingroup$ ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)... $\endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04
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To offer a slightly different perspective, due to the Spectral Theorem you have $$ A=P_1+4P_2,$$ where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus $$ P_1=\tfrac12\,\begin{bmatrix} 1&1\end{bmatrix}\begin{bmatrix} 1\\1\end{bmatrix}=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $$ And then $$ A=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}+4\begin{bmatrix}1/2&-1/2\\-1/2&1/2\end{bmatrix}=\begin{bmatrix}5/2&-3/2\\-3/2&5/2 \end{bmatrix} $$

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Let :

$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

Since the eigenvalues are $\lambda =1 $ and $\lambda = 4$, it must be

$$\det(A-\lambda I) = \begin{vmatrix} a - \lambda & b \\ c & d - \lambda\end{vmatrix} = (a-\lambda)(d-\lambda)-bc $$

such that $\lambda = 1$ and $\lambda = 4$ are solutions to the equation :

$$(a-\lambda)(d-\lambda)-bc=0$$

Since $(1,1)^\mathbf{T}$ is an eigenvector of $A$ for $\lambda = 1$, it is :

$$(A- I)(1,1)^\mathbf{T} = 0 \Rightarrow \begin{pmatrix} a - 1 & b \\ c & d - 1\end{pmatrix}\begin{pmatrix}1 \\ 1 \end{pmatrix} =\begin{pmatrix}0 \\ 0 \end{pmatrix} $$

$$\Leftrightarrow$$

$$\begin{cases} a-1 + b = 0 \\c + d-1 = 0\end{cases}$$

Can you now find $a,b,c$ and $d$ ?

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  • $\begingroup$ Also use the other two conditions: $ \lambda_1 \lambda_2=4=ad-bc$ and $ \lambda_1+\lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, \ ad-bc=4$. $\endgroup$
    – Why
    Dec 7 '18 at 22:10

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