# Dimension of Null Spaces of Similar Matrices are the Same

## Problem 222

Suppose that $n\times n$ matrices $A$ and $B$ are similar.

Then show that the nullity of $A$ is equal to the nullity of $B$.

In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of $B$.

Add to solve later

Sponsored Links

Contents

## Definitions/Hint.

The **null space (kernel)** of an $m \times n$ matrix $A$ is the subspace of $\R^m$ defined by

\[\calN(A)=\{\mathbf{x} \in \R^m \mid A\mathbf{x}=\mathbf{0}\}.\]

Two matrices $A$ and $B$ are **similar** if there exists an invertible matrix $S$ such that

\[A=S^{-1}BS.\]

To show that the dimensions of $\calN(A)$ and $\calN(B)$ are equal, find an isomorphism between these vector spaces using the fact that matrices $A$ and $B$ are similar.

## Proof.

Since $A$ and $B$ are similar, there exists an invertible matrix $S$ such that

\[A=S^{-1}BS.\]
Observe that if $\mathbf{x}\in \calN(A)$, then we have

\begin{align*}

A\mathbf{x}=\mathbf{0}\\

\Leftrightarrow

(S^{-1}BS)\mathbf{x}=\mathbf{0}\\

\Leftrightarrow

B(S\mathbf{x})=\mathbf{0}.

\end{align*}

Therefore we have

\[S\mathbf{x} \in \calN(B).\]
From this observation, we define the map

\[\Psi: \calN(A) \to \calN(B)\]
by sending $\mathbf{x} \in \calN(A)$ to $S\mathbf{x}\in \calN(B)$.

We claim that the map $\Psi$ is an isomorphism of vector spaces.

To see that $\Psi$ is a linear transformation, let $\mathbf{x}, \mathbf{y} \in \calN(A)$, and $c$ be a scalar.

Then we have

\begin{align*}

\Psi(\mathbf{x}+\mathbf{y})=S(\mathbf{x}+\mathbf{y})=S\mathbf{x}+S\mathbf{y}=\Psi(\mathbf{x})+\Psi(\mathbf{y})

\end{align*}

and

\begin{align*}

\Psi(c\mathbf{x})=S(c\mathbf{x})=cS\mathbf{x}=c\Psi(\mathbf{x}).

\end{align*}

Thus $\Psi$ is a linear transformation.

To show that $\Psi$ is an isomorphism, we give the inverse linear transformation of $\Psi$.

We define $\Phi:\calN(B) \to \calN(A)$ to be a map sending $\mathbf{x} \in \calN(B)$ to $S^{-1}\mathbf{x} \in \calN(A)$.

By a similar argument as above, we can show that the element $S^{-1}\mathbf{x}$ is indeed in $\calN(A)$ and $\Phi$ is a linear transformation and it is straightforward to see that $\Psi\circ \Phi=\id_{\calN(B)}$ and $\Phi \circ \Psi=\id_{\calN(A)}$.

Hence $\Phi$ is the inverse of $\Psi$, and $\Psi$ is an isomorphism.

Therefore the vector spaces $\calN(A)$ and $\calN(B)$ are isomorphic, and hence their dimensions are the same.

## Comment.

Instead of finding the inverse linear transformation, you may directly show that the map $\Psi$ is bijective (injective and surjective).

Add to solve later

Sponsored Links