# Given Graphs of Characteristic Polynomial of Diagonalizable Matrices, Determine the Rank of Matrices

## Problem 217

Let $A, B, C$ are $2\times 2$ diagonalizable matrices.

The graphs of characteristic polynomials of $A, B, C$ are shown below. The red graph is for $A$, blue one for $B$, and green one for $C$.

From this information, determine the rank of the matrices $A, B,$ and $C$.

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## Hint.

Observe that a null space a matrix $M$ is the same as the eigenspace $E_0$ corresponding to the eigenvalue $\lambda=0$ (if 0 is an eigenvalue).

So what you need to see is whether the graphs pass though the origin.

Also, since the matrices are diagonalizable, the algebraic multiplicities are the same as the geometric multiplicities (the dimension of the eigenspace).

You can determine the algebraic multiplicities from the graph by looking at whether the graphs are tangential to $x$-axis or not.

## Solution.

We first determine the nullities of the matrices, and then using the rank-nullity theorem we obtain the ranks.

In general the nullity of any matrix $M$ is the dimension of the null space

\[\calN(M)=\{\mathbf{x}\in \R^2 \mid M\mathbf{x}=\mathbf{0}\}\]
and the null space is the same as the eigenspace

\[E_{0}=\{\mathbf{x}\in \R^2\mid M\mathbf{x}=0\mathbf{x}=\mathbf{0}\}\]
corresponding to the eigenvalue $\lambda=0$ (if any).

From this observation, we see that the nullity of a matrix $M$ is the geometric multiplicity of the eigenspace $E_0$ associated to the eigenvalue $0$ (if any).

### The Nullity of the Matrix $A$

Now we determine the nullity of the matrix $A$.

The graph of the characteristic polynomial $p_A(\lambda)$ of $A$ passes through the origin $(0,0)$.

Thus $\lambda=0$ is a root of $p_A(\lambda)$ and hence $\lambda=0$ is an eigenvalue.

Since the $x$-axis is tangential to the graph of $p_A(\lambda)$, the algebraic multiplicity of $\lambda=0$ is $2$. Since the matrix $A$ is diagonalizable, the geometric multiplicity is the same as the algebraic multiplicity. Therefore the nullity of $A$ is $2$.

### The Nullity of the Matrix $B$

Next, the graph of characteristic polynomial $p_B(\lambda)$ of $B$ does not pass through the origin $(0,0)$. Thus $\lambda=0$ is not an eigenvalue of $B$. This yields that the matrix $B=B-0I$ is nonsingular matrix, and hence we have the null space $\calN(B)=\{0\}$ and the nullity of $B$ is zero.

### The Nullity of the Matrix $C$

Third, the origin $(0,0)$ is on the graphs of characteristic polynomial $p_C(\lambda)$ of $C$, but the $x$-axis is not tangential to the graph of $p_C(\lambda)$.

Therefore the algebraic (hence geometric) multiplicity of $\lambda=0$ is $1$. Thus the nullity of $C$ is $1$.

### Ranks by the Rank-Nullity Theorem

Finally, using the rank-nullity theorem

\[\text{rank}+\text{nullity}=2,\]
we obtain the rank of $A, B, C$ are $0, 2, 1$, respectively.

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