# Finding the eigenvalues

Given a $3 \times 3$ matrix with real entries such that $\det (A) = 6$ and $\mathrm{trace}( A )= 0$.

If $\det (A+I)=0$ where $I$ is a $3 \times 3$ identity matrix , then eigenvalues of $A$ are $-1$, $-2$, $3$.

I am confused. Can we conclude from the given data that $-1$ is one of the eigenvalues. Rest follows as trace = sum of eigenvalues and determinant = product of eigenvalues.

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Again: in English, it's "eigenvalues", single word, not "eigen values". – Arturo Magidin May 17 '12 at 16:31

$\lambda$ is an eigenvalue of $A$ if and only if $\lambda + \mu$ is an eigenvalue of $A+\mu I$. To see this, note that $A\mathbf{v}=\lambda\mathbf{v}$ if and only if $A\mathbf{v}+\mu\mathbf{v}=(\lambda+\mu)\mathbf{v}$, if and only if $(A+\mu I)\mathbf{v}=(\lambda+\mu)\mathbf{v}$.

The fact that the determinant of $A+I$ is $0$ tells you that $0$ is an eigenvalue of $A+I$, and therefore, that $-1$ is an eigenvalue of $A$ (so that $-1+1=0$ will be an eigenvalue of $A+I$).

Once you know that $-1$ is an eigenvalue of $A$, you know that the other two eigenvalues add up to $1$ (so the trace will be $0$), and have a product of $-6$ (so the determinant will be $6$). So you are looking for $r_1$ and $r_2$ with $r_1+r_2=1$, $r_1r_2=-6$. This gives you the quadratic $$(\lambda-r_1)(\lambda-r_2) = \lambda^2 -\lambda - 6$$ whose roots are $3$ and $-2$, giving you the other two eigenvalues.

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Many thanks for detailed reply. You explain it so well. – preeti May 17 '12 at 17:02

Since $\det(A+I) = 0$, the homogeneous system of equations $(A+I)x = 0$ has a non-trivial solution, i.e. $Ax = -Ix = -1x$ for some non-zero vector $x$.

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This is built into the definition of eigenvalue. We have $\lambda$ is an eigenvalue of $A$ iff there is a non-zero $v$ such that $Av=\lambda v$, meaning that $(A-\lambda I)v=0$. That is the case precisely if $A-\lambda I$ is not invertible, which holds precisely if the determinant of $A-\lambda I$ is $0$. Put $\lambda=-1$.

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Thanks a lot for the help. – preeti May 17 '12 at 17:06