Determine a Matrix From Its Eigenvalue

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 259

Let
\[A=\begin{bmatrix}
a & -1\\
1& 4
\end{bmatrix}\] be a $2\times 2$ matrix, where $a$ is some real number.
Suppose that the matrix $A$ has an eigenvalue $3$.

(a) Determine the value of $a$.

(b) Does the matrix $A$ have eigenvalues other than $3$?

 
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Solution.

(a) Determine the value of $a$

Since $3$ is an eigenvalue of the matrix $A$, we have
\[0=\det(A-3I),\] where $I$ is the $2\times 2$ identity matrix.
Thus we have
\begin{align*}
0&=\det(A-3I)\\
&=\begin{vmatrix}
a-3 & -1\\
1& 4-3
\end{vmatrix}\\
&=\begin{vmatrix}
a-3 & -1\\
1& 1
\end{vmatrix}\\
&=(a-3)(1)-(-1)(1)=a-2.
\end{align*}
Thus the value of $a$ must be $2$.

(b) Does the matrix $A$ have eigenvalues other than $3$?

Let us determine all the eigenvalue of the matrix
\[A=\begin{bmatrix}
2 & -1\\
1& 4
\end{bmatrix}.\]

We compute the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
2-t & -1\\
1& 4-t
\end{vmatrix}\\
&=(2-t)(4-t)-(-1)(1)\\
&=t^2-6t+9\\
&=(t-3)^2.
\end{align*}

Since the eigenvalues are roots of the characteristic polynomial, solving $(t-3)^2=0$ we see that $t=3$ is the only eigenvalue of $A$ (with algebraic multiplicity $2$).
Hence the matrix $A$ does not have eigenvalues other than $3$.


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