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Given the following matrix: $\begin{bmatrix}2 & 1\\2 &3\end{bmatrix}$ Find the Eigenvalues and Eigenvectors. So I found the Eigenvalues to be 2 and 3. Then I plugged in 2 to find the corresponding eigenvector and got through the homogenous augmented matrix and got $2x_1+x_2=0$ which means $x_1=0$, and I also got $x_2=0$. So is the Eigenvector just $\begin{bmatrix}0 \\0\end{bmatrix}$ Then I tried for find the eigenvector for $3$ and through the augmented homogeneous matrix I got $x_1=x_2$ and $2x_1=-x_2$ so I got this eigenvector to be $\begin{bmatrix}-1 \\2\end{bmatrix}$.

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    $\begingroup$ An eigenvector, by definition, is non-zero. And the eigenvalues you found are false. How did you find them? $\endgroup$ – Bernard Apr 27 '16 at 0:42
  • $\begingroup$ I believe you'll find that your eigenvalues are incorrect: check your characteristic polynomial again (you should have $ \ \lambda^2 \ - \ 5 \lambda \ + \ 4 \ = \ 0 \ $ ) . $\endgroup$ – colormegone Apr 27 '16 at 0:42
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Let $A=\begin{bmatrix}2 & 1\\2 &3\end{bmatrix}$. The roots of the characteristic polynomial for the matrix coincide with its eigenvalues, so we first factor the polynomial: $$p_A(t)=det(tI-A)=det\begin{bmatrix}t-2 & -1\\-2 &t-3\end{bmatrix}=(t-2)(t-3)-(-1)(-2)=t^2-5t+6-2=t^2-5t+4=(t-4)(t-1)$$ Thus, the roots of $p_A(t)$ are $4$ and $1$, which are the eigenvalues of the matrix. Now to find the $\lambda_1=4$ eigenspace we consider the kernel of the matrix $4I-A$, which is spanned by the vector $v_1=\begin{bmatrix}1\\2\end{bmatrix}$. Similarly, the $\lambda_2=1$ eigenspace of A is the kernel of $I-A$, which is spanned by the vector $v_2=\begin{bmatrix}-1\\1\end{bmatrix}$. Therefore, the eigenvalues of $A$ are $4$ and $1$, with corresponding eigenvectors $v_1$ and $v_2$, respectively.

Also, note that if you are finding eigenvectors of a matrix, and you obtain the zero vector, something has gone wrong and you should retry your calculations, because an eigenvector is by definition nonzero.

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