Question

Let A be a 2 x 2 matrix, and suppose that $A^2$ = 0. Then for each scalar c, det(cI - A) = $c^x$. Find x.

Answer 1

2

Answer 2

To solve this problem, we will use properties of 2x2 matrices, including the characteristic polynomial, trace, determinant, and the Cayley-Hamilton theorem.

Let A be a 2x2 matrix. Its characteristic polynomial is given by: $P(\lambda) = \det(A - \lambda I) = \lambda^2 - \text{tr}(A)\lambda + \det(A)$

We are asked to find $x$ such that $\det(cI - A) = c^x$. Note that $\det(cI - A) = \det(-(A - cI)) = (-1)^2 \det(A - cI) = \det(A - cI)$. So, $\det(cI - A)$ is simply the characteristic polynomial evaluated at $\lambda = c$: $\det(cI - A) = c^2 - \text{tr}(A)c + \det(A)$

Now, let's use the given condition $A^2 = 0$. According to the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. For a 2x2 matrix A, this means: $A^2 - \text{tr}(A)A + \det(A)I = 0$

Substitute the given condition $A^2 = 0$ into this equation: $0 - \text{tr}(A)A + \det(A)I = 0$ $\text{tr}(A)A = \det(A)I$

We need to consider two cases for matrix A:

Case 1: A is the zero matrix. If $A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, then $A^2 = 0$ is trivially true. In this case, $\text{tr}(A) = 0 + 0 = 0$ and $\det(A) = (0)(0) - (0)(0) = 0$. Substitute these values into the expression for $\det(cI - A)$: $\det(cI - A) = c^2 - (0)c + 0 = c^2$ Comparing this with $\det(cI - A) = c^x$, we get $c^x = c^2$, which implies $x=2$.

Case 2: A is a non-zero matrix. From the equation $\text{tr}(A)A = \det(A)I$: If $\text{tr}(A) \neq 0$, then we can write $A = \frac{\det(A)}{\text{tr}(A)}I$. Let $k = \frac{\det(A)}{\text{tr}(A)}$. So, $A = kI = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$. Now, calculate $A^2$: $A^2 = (kI)^2 = k^2I = \begin{pmatrix} k^2 & 0 \\ 0 & k^2 \end{pmatrix}$. Given that $A^2 = 0$, we have $k^2I = 0$. Since $I \neq 0$, it must be that $k^2 = 0$, which implies $k=0$. If $k=0$, then $A = 0 \cdot I = 0$, which contradicts our assumption that A is a non-zero matrix. Therefore, if A is a non-zero matrix and $A^2=0$, it must be that $\text{tr}(A)=0$.

Now, if $\text{tr}(A)=0$, the equation $\text{tr}(A)A = \det(A)I$ becomes: $0 \cdot A = \det(A)I$ $0 = \det(A)I$ Since $I \neq 0$, it must be that $\det(A)=0$.

So, for any 2x2 matrix A such that $A^2=0$ (whether A is zero or non-zero), we must have $\text{tr}(A)=0$ and $\det(A)=0$.

Substitute these values into the expression for $\det(cI - A)$: $\det(cI - A) = c^2 - \text{tr}(A)c + \det(A)$ $\det(cI - A) = c^2 - (0)c + 0$ $\det(cI - A) = c^2$

Given that $\det(cI - A) = c^x$. Comparing the two expressions, we get $c^x = c^2$. Therefore, $x=2$.