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Could you explain to me how to solve matrix equations?

Here is an example:

Prove that:

$$2X^2 + X = \begin{bmatrix} -1&5&3\\-2&1&2\\0&-4&-3\end{bmatrix}$$

has no solutions in $M(3,3;\mathbb{R})$, where $M(3,3;\mathbb{R})$ is the space of all $3\times3$ matrices with real entries.

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For clarity, do you mean $X$ instead of $x$ to represent a matrix as opposed to a variable? Regards –  Amzoti Jan 10 '13 at 18:46
    
Yes, I meant a matrix. It's already corrected. –  Bilbo Jan 10 '13 at 19:06

2 Answers 2

First note that $X$ is a $3\times 3$ real matrix and hence it must have at least one real eigenvalue.

Consider the characteristic polynomial of $$A=\begin{pmatrix} -1&5&3\\-2&1&2\\0&-4&-3\end{pmatrix}$$ which is $$p(\lambda) = -\lambda^3 -3\lambda^2 -17\lambda -11$$ Taking $A=2X^2 + X$ in $p$ gives $$p(A) = -8X^6 - 12X^5 -18X^4 -13X^3 - 37X^2 - 17X - 11 = 0$$ Note that in particular, $$q(x) = -8x^6 - 12x^5 -18x^4 -13x^3 - 37x^2 - 17x - 11 = 0$$ is an annihilating polynomial of $X$. Therefore the eigenvalues of $X$ must be amongst the roots of $q$. But all of $q$'s roots are complex, a contradiction.

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Could you explain to me why it wrong to assume matrices A, B and C as in the first answer and then show that matrix $b^2 - 4AC$ has a negative determinant? –  Bilbo Jan 11 '13 at 6:30
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The answer is a bit unclear but I don't think checking the determinant is what is intended. Rather Elias was trying to find the square root of the matrix (to solve the quadratic equation), which isn't always well defined. –  EuYu Jan 11 '13 at 15:57
    
Aha. But if I proved that the above mentioned matrix has negative determinant, it would give me the needed contradiction, wouldn't it? –  Bilbo Jan 11 '13 at 16:13
    
You first have to note that $A,\ B$ and $C$ defined in Elias' answer all necessarily commute with $X$. That is crucial to the argument. With this in mind, then yes, proving that the "discriminant" has a negative determinant will show that there are no solutions. –  EuYu Jan 11 '13 at 19:41

Since $A = \pmatrix{-1 & 5 & 3\cr -2 & 1 & 2\cr 0 & -4 & 3\cr}$ is equal to a polynomial is $X$, it commutes with $X$. Now $0 = 2 X^2+ X - A = 2 (X + I/4)^2 - I/8 - A$.
Now $A$ has a simple eigenvalue $\lambda \approx -.715923208$. If $v$ is an eigenvector of $A$ for this eigenvalue, it is also an eigenvector of $X$ for eigenvalue $\mu$, where $2 (\mu + 1/4)^2 = 1/8 + \lambda$. But since $1/8 + \lambda < 0$, $\mu$ can't be real. Now $v$ can be taken to be real (since $A$ and $\lambda$ are real), so $X v = \mu v$ is impossible if $X$ is real.

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