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- Feb 13, 2012

- 1,704

Is...Find all square matricx $A$ that satisfy the equation $A^2-4A+3I=0$

where $I$ is a Identity matrix

$\displaystyle A^{2}- 4 A +3 I= A\ (A -4\ I + 3\ A^{-1})= A\ (A -3\ I - I + 3\ A^{-1})= A\ \{(A-3\ I)- A^{-1}\ (A-3\ I)\}= A\ (A-3\ I)\ (I-A^{-1})$

... so that the A satisfying the given equation also satisfy one of the equations...

$\displaystyle A-3\ I=0$

$\displaystyle A-I=0$

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

The quadratic in A can be factorised straight off without having to worry about the existance of inverses. But I'm not sure that we can conclude that if a product is zero then one of the terms in the product is zero in this case.Is...

$\displaystyle A^{2}- 4 A +3 I= A\ (A -4\ I + 3\ A^{-1})= A\ (A -3\ I - I + 3\ A^{-1})= A\ \{(A-3\ I)- A^{-1}\ (A-3\ I)\}= A\ (A-3\ I)\ (I-A^{-1})$

... so that the A satisfying the given equation also satisfy one of the equations...

$\displaystyle A-3\ I=0$

$\displaystyle A-I=0$

Kind regards

$\chi$ $\sigma$

CB

- Feb 13, 2012

- 1,704

A trivial test confirms that $A=I$ and $A=3\ I$ are solutions of the equation...The quadratic in A can be factorised straight off without having to worry about the existance of inverses. But I'm not sure that we can conclude that if a product is zero then one of the terms in the product is zero in this case.

CB

$\displaystyle A^{2}-4\ A+3\ I=0$ (1)

The question is: there are some other solutions to (1)?... it is clear that the question is strictly connected to another question: can exist two matrices X and Y, with $X\ne 0$ and $Y\ne 0$, for whose is $X\ Y=0$?... a very interesting question!...

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

The answer to the last question is yes:A trivial test confirms that $A=I$ and $A=3\ I$ are solutions of the equation...

$\displaystyle A^{2}-4\ A+3\ I=0$ (1)

The question is: there are some other solutions to (1)?... it is clear that the question is strictly connected to another question: can exist two matrices X and Y, with $X\ne 0$ and $Y\ne 0$, for whose is $X\ Y=0$?... a very interesting question!...

Kind regards

$\chi$ $\sigma$

\[ A=\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] \]

Then \( A^2=0 \)

CB

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Yes.Thanks to all experts

but my doubt is , is there is any other matrix exists for which $A=\begin{bmatrix}a &b \\

c & d

\end{bmatrix}$ which satisfy $A^2-4A+3I=0$

$$

A = \begin{bmatrix}3&0\\0&1\end{bmatrix}

$$

I just used the Cayley-Hamilton Theorem in reverse.

- Jan 26, 2012

- 890

Which is that a matrix \(A\) satisfies its' own characteristics equation:Yes.

$$

A = \begin{bmatrix}3&0\\0&1\end{bmatrix}

$$

I just used the Cayley-Hamilton Theorem in reverse.

\[\det (A-\lambda I)=0\]

Which off course allows you to generate many 2x2 solutions of the original equation.

But is that (with the two we already have, which are not restricted to being 2x2) all of them?

CB

- Jan 26, 2012

- 890

OK, I know it is bad form to reply to ones own posts, but:Which is that a matrix \(A\) satisfies its' own characteristics equation:

\[\det (A-\lambda I)=0\]

Which off course allows you to generate many 2x2 solutions of the original equation.

But is that (with the two we already have, which are not restricted to being 2x2) all of them?

CB

Let us restrict our attention to 2x2 matrices. We know that \(A\) satisfies its' charateristic equation, which we may write as: \(\lambda^2+b\lambda+c=0\) and so \(A^2+bA+cI=0\), now suppose it also satisfies another second degree equation: \(A^2+\beta A+\gamma I =0\) (which forces \(\beta \ne b\) if these are to be different and consistent).

Then we have: \((b-\beta)A+(c-\gamma)I=0\), and so: \(A=\frac{\gamma-c}{b-\beta}I\)

So the 2x2 solutions to \(A^2-4A+3I=0 \) are either matrices with charateristic equation \(\lambda^2-4\lambda+3=0\) or are multiples of the identity (and so of neccessity the solutions we found by factorising the equation)

CB

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