Maximize the Dimension of the Null Space of $A-aI$

Ohio State University exam problems and solutions in mathematics

Problem 200

\[ A=\begin{bmatrix}
5 & 2 & -1 \\
2 &2 &2 \\
-1 & 2 & 5

Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.

Your score of this problem is equal to that dimension times five.

(The Ohio State University Linear Algebra Practice Problem)
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Note that if $a$ is not an eigenvalue of the matrix $A$, then the matrix $A-aI$ is nonsingular, that is, the null space of $A-aI$ is the zero space, and the dimension is $0$.
Thus to obtain a nonzero point, your favorite number should be an eigenvalue.

Let us find the eigenvalue. The characteristic polynomial $p(t)$ of $A$ is
5-t & 2 & -1 \\
2 &2-t &2 \\
-1 & 2 & 5-t
2-t & 2\\
2& 5-t
2 & 2\\
-1& 5-t
2 & 2-t\\
-1& 2
&\text{(By the first row cofactor expansion.)}\\

Eigenvalues are roots of the characteristic polynomial.
Thus eigenvalues of $A$ are $0$ and $6$, with algebraic multiplicities $1$ and $2$, respectively.

Since the matrix $A$ is a real symmetric matrix, it is diagonalizable. Thus the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity.

Therefore the dimension of the null space of $A-6I$, which is the geometric multiplicity of the eigenvalue $6$ is equal to $2$, and the dimension of the null space of $A$ is $1$.

Thus, if you take $a=6$, you get $2\times 5=10$ points. If your favorite number is $a=0$, then your score is $1\times 5=5$ points. The other choice of $a$ will be zero points.


If you didn’t notice that the given matrix is a real symmetric matrix, then you need to find eigenvectors corresponding to eigenvalues $6$ and to show that the eigenspace has dimension $2$.

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