Find the determinant of $I+A$

Question

Let $A$ be a $2\times2$ matrix with real entries such that $A^2=0$.Find the determinant of $I+A$ where $I$ denotes the identity matrix. I proceed in this way :Note that $(I+A)A=A+A^2 \Longrightarrow (I+A)A=A$ (Since $A^2=0$).

Now taking determinant both side we get $|(I+A)A| = |A| \Longrightarrow |(I+A)|\cdot|A|=|A|\Longrightarrow |I+A|=1$.

Am I proceed in correct way? In the last line am I need to consider $|A|=0$?

Since $A^2=0$ $A$ is not invertible, therefore $|A|=0$, so your proof is not correct. — sebigu, Aug 02 '12 at 13:25
Better consider the characteristic polynomial of $A$. You can compute the one from $A+I$ from it and read out the determinant. — sebigu, Aug 02 '12 at 13:27
As an alternative to @sebigu's hint: you can show that a nilpotent matrix is similar to a strictly triangular matrix. Now, what is the determinant of the sum of the identity and a strictly triangular matrix? — J. M. ain't a mathematician, Aug 02 '12 at 13:33

score 12 · Answer 1 · answered Aug 02 '12 at 13:32

$A$ is similar to (and hence has the same determinant as) a matrix of the form $\begin{pmatrix} 0 & * \\ 0 & 0 \end{pmatrix}$. That is, there is an invertible matrix $P$ such that

$$A':=P^{-1}AP = \begin{pmatrix} 0 & * \\ 0 & 0 \end{pmatrix}$$

and $\det A = \det A'$. But $I':=P^{-1}IP=I$, and so $P^{-1}(I+A)P=I+A'$, and so $\det (I+A) = \det (I+A')$.

So what is $\det (I+A)$?

score 6 · Accepted Answer · answered Aug 02 '12 at 13:57

6

The matrix $A$ is nilpotent with characteristic polynomial $x^2.$ But by definition $x^2 = \det(A - xI).$ If we let $x = -1,$ then we have $\det(A + I) = 1.$

answered Aug 02 '12 at 13:57

This is the simplest way. Than you for help – Argha Aug 02 '12 at 14:16

score 5 · Answer 3 · answered Aug 02 '12 at 13:45

5

Let $x = \det (I+A)$

$$(I+A).(I+nA)=I+(n+1)A$$

Hence $\det(I+nA)=x^n$

But $\det(I+nA)=P(n)$ with P a polynomial (of maximum degree the size of matrix A).

So the only solutions are $x=1$ or $x=0$. But If $x=0$, there is a $v\neq 0$ such that $v+Av=0$, hence, $Av=-v$, and $A^2v=v$, that is not possible.

So $\det(I+A)=1$

Note that this proof is true for any size of matrix.

answered Aug 02 '12 at 13:45

Xoff

10,310

Very nice answer, +1 – Clive Newstead Aug 02 '12 at 14:01

score 2 · Answer 4 · answered Aug 02 '12 at 13:35

2

I like this one:

The characteristic polynomial of $A$ is $x^2$. By substitution we get $$(x-1)^2=x^2-2x+1$$ as characteristic polynomial of $A+I$. Now we see trace and determinant.

answered Aug 02 '12 at 13:35

sebigu

777
3
7

score 2 · Answer 5 · edited Mar 23 '13 at 13:58

The trace of $A$ is $0$, using $$A^2-\operatorname{tr}(A)\cdot A+(\det A)\cdot I=0$$ and the fact that the determinant of $A$ is $0$. Applying this to $A+I$, we get $$\tag{1} A+2I-\operatorname{tr}(A+I)\cdot(A+I)+\det(A+I)\cdot I=0,$$ hence $$2A+I-2(A+I)+\det(A+I)\cdot I=0,$$ and $$\det(A+I)\cdot I=I,$$ hence $\det (A+I)$.

The formula $(1)$ can we deduce either by Cayley Hamilton theorem (sledgehammer), or by hand.

Find the determinant of $I+A$

5 Answers5