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Thread: Show that A must be invertible

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    Show that A must be invertible

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

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

    $\displaystyle a_0I_m=-(a_1A+\cdots +a_nA^n)\,\Longrightarrow I_m=-\frac{a_1}{a_0}A-\frac{a_2}{a_0}A^2-\cdots -\frac{a_n}{a_0}A^n$ , 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? :

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

    Tonio
    ok, after factor I got $\displaystyle Im = A (-a1/a0 - a2/a0A - ....an/a0A^n-1)$
    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 $\displaystyle Im = A (-a1/a0 - a2/a0A - ....an/a0A^n-1)$
    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 $\displaystyle BA=AB=I$ 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 $\displaystyle BA=AB=I$ for some square matrix B...right? Well, you've just found B above...
    oh, awesome, I got it, thanks alot
    Last edited by 450081592; Nov 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 $\displaystyle A^2$, if $\displaystyle P(x) = x^2 +x -2$, calculate P(A). Calculate $\displaystyle A^-1$.) Can you explan this,>?
    Last edited by 450081592; Nov 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 $\displaystyle A^2$, if $\displaystyle P(x) = x^2 +x -2$, calculate P(A). Calculate $\displaystyle A^-1$.) Can you explan this,>?
    The first matrix is what's called a scalar matrix, in this case $\displaystyle 2\cdot I_3=2\!\cdot\!\left(\begin{array}{ccc}1&0&0\\0&1&0 \\0&0&1\end{array}\right)=\left(\begin{array}{ccc} 2&0&0\\0&2&0\\0&0&2\end{array}\right)$ , and this kind of matrices is what one must use when doing matrix equations:

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

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

    Tonio
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