# Show that A must be invertible

• November 27th 2009, 02:42 PM
450081592
Show that A must be invertible
Let $P(x) = a0 + a1x + a2x^2 + ... + anx^n$ be a polynomial of degree n in x with real coefficients. For any mxm matrix A, we define $P(A) = a0Im + a1A + a2A^2 + ... + anA^n$.
Show that, if $P(A) = 0m$ and $a0 not = 0$, then A must be invertible. [Hint: isolate Im from the equation P(A) = 0m]
• November 27th 2009, 03:18 PM
tonio
Quote:

Originally Posted by 450081592
Let $P(x) = a0 + a1x + a2x^2 + ... + anx^n$ be a polynomial of degree n in x with real coefficients. For any mxm matrix A, we define $P(A) = a0Im + a1A + a2A^2 + ... + anA^n$.
Show that, if $P(A) = 0m$ and $a0 not = 0$, then A must be invertible. [Hint: isolate Im from the equation P(A) = 0m]

Why don't you do what the hints says?(Angry) :

$a_0I_m=-(a_1A+\cdots +a_nA^n)\,\Longrightarrow I_m=-\frac{a_1}{a_0}A-\frac{a_2}{a_0}A^2-\cdots -\frac{a_n}{a_0}A^n$ , and now just factor out A in the right side...

Tonio
• November 27th 2009, 03:31 PM
450081592
Quote:

Originally Posted by tonio
Why don't you do what the hints says?(Angry) :

$a_0I_m=-(a_1A+\cdots +a_nA^n)\,\Longrightarrow I_m=-\frac{a_1}{a_0}A-\frac{a_2}{a_0}A^2-\cdots -\frac{a_n}{a_0}A^n$ , and now just factor out A in the right side...

Tonio

ok, after factor I got $Im = A (-a1/a0 - a2/a0A - ....an/a0A^n-1)$
what does this imply, does it mean A times a elementary matrix equal to Im which is the identity, so A in invertible?
• November 27th 2009, 03:38 PM
tonio
Quote:

Originally Posted by 450081592
ok, after factor I got $Im = A (-a1/a0 - a2/a0A - ....an/a0A^n-1)$
what does this imply, does it mean A times a elementary matrix equal to Im which is the identity, so A in invertible?

You know a square matrix A is invertible iff $BA=AB=I$ for some square matrix B...right? Well, you've just found B above...
• November 27th 2009, 04:14 PM
450081592
Quote:

Originally Posted by tonio
You know a square matrix A is invertible iff $BA=AB=I$ for some square matrix B...right? Well, you've just found B above...

oh, awesome, I got it, thanks alot(Rofl)
• November 27th 2009, 04:51 PM
450081592
by the way what is
[2 0 0]
[0 2 0]
[0 0 2] -2 ? it's like the identity but it's 2 instead of 1, can the matrix subtract a number

For A =
[2 2 -1]
[-1 -1 1]
[2 4 -1] , compute $A^2$, if $P(x) = x^2 +x -2$, calculate P(A). Calculate $A^-1$.) Can you explan this,>?
• November 28th 2009, 01:19 AM
tonio
Quote:

Originally Posted by 450081592
by the way what is
[2 0 0]
[0 2 0]
[0 0 2] -2 ? it's like the identity but it's 2 instead of 1, can the matrix subtract a number

For A =
[2 2 -1]
[-1 -1 1]
[2 4 -1] , compute $A^2$, if $P(x) = x^2 +x -2$, calculate P(A). Calculate $A^-1$.) Can you explan this,>?

The first matrix is what's called a scalar matrix, in this case $2\cdot I_3=2\!\cdot\!\left(\begin{array}{ccc}1&0&0\\0&1&0 \\0&0&1\end{array}\right)=\left(\begin{array}{ccc} 2&0&0\\0&2&0\\0&0&2\end{array}\right)$ , and this kind of matrices is what one must use when doing matrix equations:

$P(A)=A^2+A-2\!\!\cdot\!\!I=\left(\begin{array}{rrr}2&2&-1\\-1&-1&1\\2&4&-1\end{array}\right)\left(\begin{array}{rrr}2&2&-1\\-1&-1&1\\2&4&-1\end{array}\right)+$ $\left(\begin{array}{rrr}2&2&-1\\-1&-1&1\\2&4&-1\end{array}\right)-\left(\begin{array}{rrr}2&0&0\\0&2&0\\0&0&2\end{ar ray}\right)$

Evaluate the above equation (you must get the zero matrix), and then from this calculate $A^{-1}$ ...

Tonio