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Thread: A Problem with a Norm of a Matrix

  1. #1
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    A Problem with a Norm of a Matrix

    Hi,
    I am working on a Linear Algebra problem set and am running into a problem (which is probably basic, I just forget how the operator works).

    Here is the question:#1 Show that norm of the following expression is equal to 1. A Problem with a Norm of a Matrix-problem-1.gif
    Where the Capital X denotes a matrix.

    What I am confused on is how to deal with the matrix which is raised to the exponent 1/2. Do I treat this as a regular exponent? Or, is there another way to treat these exponents for matrices?
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  2. #2
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    Re: A Problem with a Norm of a Matrix

    I think there is something wrong with what you have written because expressions like $\displaystyle X^T X$ usually work for vector norms. Using a square matrix of any size $\displaystyle 2\times 2$ will give you a vector back. Indeed, if $\displaystyle X^T = \begin{bmatrix}x_1 \dots x_n\end{bmatrix}$, then

    $\displaystyle X^T X = X\cdot X = x_1^2+\dots x_n^2$

    and taking the square root gives you the norm of the vector. Since you only have one index in the sum, I would assume you are talking about vectors $\displaystyle X$. In that case you simply have that $\displaystyle X/||X||$ is a unit vector.

    More generally, if $\displaystyle ||\cdot||$ is some norm (vector or matrix, if doesn't matter), then you have

    $\displaystyle ||kA|| = |k| ||A||$

    Since $\displaystyle ||X||$ is a constant, you have

    $\displaystyle \left|\left|\frac{X}{||X||} \right|\right| = \frac{1}{||X||}||X|| = 1$

    You should check exactly what the question is asking you.

    Now, if you are interested in the square roots of matrices, they are not numbers. If $\displaystyle A$ is a square matrix, we seek to find a matrix $\displaystyle B$ such that $\displaystyle B^2 = A$ and we call $\displaystyle B = \sqrt{A}$. In general, $\displaystyle B$ is not unique. You may read more about it here Square root of a matrix - Wikipedia, the free encyclopedia
    Last edited by Vlasev; Aug 26th 2012 at 08:00 PM.
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