# Proof that Nullity of similar matrices is the same

Problem:

Suppose the matrix $\mathbf A$ is similar to $\mathbf B$, with $\mathbf B=\mathbf S^{-1}\mathbf A\mathbf S$.

a. Show that if $\mathbf x\in\mathrm{ker}(\mathbf B)$, then $\mathbf {Sx}\in \mathrm{ker}(\mathbf A).$

b.Show that $\mathrm{nullity}(\mathbf A)=\mathrm{nullity}(\mathbf B)$. Hint: If $\{\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_p\}$ is a basis for $\mathrm{ker}(\mathbf B)$, then the vectors $\{\mathbf{Sv_1}, \mathbf{Sv_2},\ldots, \mathbf{Sv_p}\}\subset\mathrm{ker}(A)$ are linearly independent. Now reverse the roles of $\mathbf A$ and $\mathbf B$.

Solution:

a. Note that $\mathbf{AS}=\mathbf{SB}$. If $\mathbf x\in\mathrm{ker}(B)$, then $\mathbf A(\mathbf{Sx})= \mathbf{SBx}=\mathbf{S0} = \mathbf 0$, so that $\mathbf {Sx}\in\mathrm{ker}(\mathbf A)$, as claimed.

b.We use the hint and observe that $$\mathrm{nullity}(\mathbf B)=\dim(\mathrm{ker}\ \mathbf B)=p\leqslant\dim(\mathrm{ker}\ \mathbf A)=\mathrm{nullity}(\mathbf A),$$ since $\mathbf {Sv_1},\ldots,\mathbf{Sv_p}$ are $p$ linearly independent vectors in $\mathrm{ker}(\mathbf A)$. Reversing the roles of $\mathbf A$ and $\mathbf B$ shows that, conversely, $\mathrm{nullity}(\mathbf A)\leqslant \mathrm{nullity}(\mathbf B)$, so that the equation $\mathrm{nullity}(\mathbf A)= \mathrm{nullity}(\mathbf B)$ holds, as claimed.

Why can we reverse the roles of $\mathbf A$ and $\mathbf B$ in this case? The fact that vectors in $\mathrm{ker}(\mathbf A)$ are linearly independent doesn't mean they are bases of the kernel.

Thank you so much.

• We can reverse the roles of A and B because they are similar to one another therefore the argument applies both ways – Klint Qinami Mar 30 '16 at 1:40

You simply show that all of the vectors that form a basis for the kernel of B must also be a part of the basis for the kernel of A based on the same logic from 71a. More formally, $dim(ker(b)) \leq dim(ker(A))$. Additionally, you can also view the equation as $SBS^{-1} = A$ and make the same argument for $dim(ker(A)) \leq dim(ker(B))$ Since both of these hold, $dim(ker(A)) = dim(ker(B))$

• Ohhhh, got it :) Thank you so much. I should have made connection with the an a) problem. – Li Cooper Mar 30 '16 at 3:24
• No problem glad to help – Klint Qinami Mar 30 '16 at 3:25

$B = S^{-1}AS$

$SB = AS$

If $\mathbf u$ is not in the kernel of B, then $\mathbf u$ is not in the kernel of $SB$. That is, when we multiply $B\mathbf u$ we get a vector $\mathbf v$. And, since $S$ is non-singular (i.e. the vectors in $S$ are linearly independent) $S\mathbf v$ is not $0.$

If $\mathbf x$ is in the kernel of $B$, then it is in the kernel of $SB.$ And, hence it is in the kernel of $AS$

And, if $\mathbf u$ is not in the kernel of $B$, then it is not in the kernel of $AS$.

Finally if $\mathbf x_1, \mathbf x_2, \mathbf x_2$ form a basis for the kernel of $B$ then $S\mathbf x_1, S\mathbf x_2, S\mathbf x_2$ form a basis for the kernel of A.

• Thank you so much. That makes sense. I wish I could choose both answers as best. – Li Cooper Mar 30 '16 at 3:26