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

[Boom101]

Matrix Properties

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

\left( \begin{array}{ccc}<br /> -3 &amp; 0 &amp; 0\\<br /> 0 &amp; -3 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0<br /> \end{array} \right)

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]

Hello, I am having difficulties with this question. A is a 3x3 matrix, the first A is squared, and I3 is the 3x3 identity 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

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.

(does 9I3 = 9, since 9 is a constant?)

No, 9I₃ does not equal 9. 9I₃ is a diagonal matrix with 9's down the main diagonal.

, then A(A+3) + 3(A+3) = 0 but I have no idea what to do after.

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

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!