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I am taking a course in linear algebra, and one of the questions is as follows (not a quote from the questions):

The matrix $A = \begin{bmatrix} 1 & 3 \\ 4 & 2\end{bmatrix}$ has eigenvalues $5$ and $-2$, this need not be proven. Find the rank of the matrix $xI - A = \begin{bmatrix} x-1 & -3 \\ -4 & x-2\end{bmatrix}$ for each real number $x$.

I have read other questions such as this one which use the rank-nullity theorem. However, this concept has not been introduced yet, which I assume means that it should not be used to solve the problem. Instead, I have only learnt of eigenvalues, eigenvectors, diagonalisation, as well as matrix algebra. The textbook that is being used is Linear Algebra with Applications by W. Keith Nicholson and is open source, available online here. This question is asked after having completed chapter 3.3 of the textbook.

I have tried manipulating $A$ into reduced row-echelon form, but I can't seem to be able to do this for all numbers $x$, and I don't quite understand the relationship between ranks and eigenvalues yet.

ettolrach
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  • The rank of $xI-A$ is at least $1$ because the matrix has at least a non-zero element. The rank is $2$ iff $\det(xI-A)\not=0$ that is when $x\not=5$ and $x\not=-2$ – Robert Z Nov 02 '21 at 19:10
  • I'm pretty sure you may know enough to prove the rank-nullity theorem. The proof is quite elementary: Suppose $A \colon \mathbb{R}^n \to \mathbb{R}^m$ is linear. Find a subspace $U$ of $\mathbb{R}^n$ such that $\mathbb{R}^n = \text{nullspace}(A) \oplus U$. Prove that $A$ is an isomorphism from $U$ to $\text{range}(A)$. – Mason Nov 02 '21 at 19:18

2 Answers2

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Recall that $x_0$ is an eigenvaluo of $A$ if and only if $det(x_0I-A)=0$.

With this in mind, $xI-A$ would have maximal rank for any $x$ not being an eigenvalue. In your case, $rank(xI-A)=2$ for $x\ne 5,-2$.

For $x=5,-2$ you know that the rank cannot be 2. But since there is always some nonnull entry in $xI-A$, the rank must be 1.

The argument for rank 2 works for any order of the matrix $A$, but the rank-1 argument only works for 2x2 matrices with a nonnull entry putside the diagonal.

Tito Eliatron
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Collect the (representative) eigenvectors in $\mathbf B$. The eigenvalues are distinct so the eigenvectors are linearly independent. Now
$\mathbf BD= A\mathbf B$
where $D$ has the eigenvalues of $5$ and $-2$ on its diagonal -- which means $D$ is invertible. So

$2=\text{rank}\big(\mathbf B\big)=\text{rank}\big(\mathbf BD\big)= \text{rank}\big(A\mathbf B\big)\leq \text{rank}\big(A\big)\leq 2 $
Thus $\text{rank}\big(A\big) = 2$

user8675309
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