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Math Help - Show that A must be invertible

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    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]
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    Quote Originally Posted by 450081592 View Post
    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? :

    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
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    Quote Originally Posted by tonio View Post
    Why don't you do what the hints says? :

    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?
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    Quote Originally Posted by 450081592 View Post
    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...
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    Quote Originally Posted by tonio View Post
    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
    Last edited by 450081592; November 27th 2009 at 04:50 PM.
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    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,>?
    Last edited by 450081592; November 27th 2009 at 09:11 PM.
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    Quote Originally Posted by 450081592 View Post
    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
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