Matrix Properties : A2 + 6A +9I3 = 0, show that A is invertible

Boom101

Hello, I am having difficulties with this question. A^2+6A+9I3 = 0. A is a 3x3 matrix. I must show that A is invertible. I am tempted to factor but this problem deals with matrices. I know this is wrong but I come to A2+3A+3A+9=0 (does 9I3 = 9, since 9 is a constant?), then A(A+3) + 3(A+3) = 0 but I have no idea what to do after.

micromass

Doesn't

[tex]\left( \begin{array}{ccc}
-3 & 0 & 0\\
0 & -3 & 0 \\
0 & 0 & 0
\end{array} \right) [/tex]

Constitute a counter-example? Or am I going senile?

Boom101

Yeah I knew what I did was wrong. In any case, I have no idea how to show that A is invertible.

micromass

Ok, ignore my above post. I WAS being senile...

What about this approach? the matrix A is similar to a triangular matrix B. This matrix B also satisfied B2+6B+9I3=0. But since B is triangular, it is easy to see that it must have -3's on it's diagonal. Thus the determinant is -9, and hence the matrix is invertible.

Mark44

At the very least, write A^2 to indicate the square of a matrix. Better yet would be A². For the latter form, click the Go Advanced button below the text input area, and then click the X² button in the menu just above the text input area, and put an exponent of 2 in between the [ sup] and [ /sup] tags.
No, 9I₃ does not equal 9. 9I₃ is a diagonal matrix with 9's down the main diagonal.A² + 6A + 9I = 0 can be factored to (A + 3I)² = 0. One thing that we can say from this is that |A + 3I| = 0.

Dick

If A is not invertible, then there is a nonzero vector x such that Ax=0. Is that compatible with A^2+6A+9I=0?

HallsofIvy

If [itex]A^2+ 6A+ 9I= 0[/itex] then [itex]A^2+ 6A= A(A+ 6I)= 9I[/itex] so that [itex]A[(A+ 6I)/9]= I[/itex].

Do see how that tells you that A is invertible?

Boom101

Thanks Ivy

abonaser

I have the same question, I don't see how the matrix is invertible, can anyone please explain?

micromass

It's inverse is (A+6I)/9

abonaser

Thank you!