Vector transpose times matrix

linear algebramatrices

my linear algebra is a bit rusty and I'm having trouble performing the following operation

$(\mu_1-\mu_0)^{T}\Sigma^{-1}(\mu_1-\mu0)$

Both $\mu_1$ and $\mu_0$ are two-dimensional vectors.
$\Sigma$ is a 2×2 matrix.

I assume this means we take the transpose of the vector difference, multiply it by the inverse of the matrix and further multiply by the vector difference.

If I am not entirely wrong this means we have to do the dot product between

$\begin{pmatrix}a \\ b\end{pmatrix} \begin{pmatrix}c & d \\ e & f \end{pmatrix} \begin{pmatrix}g & h \end{pmatrix}$

As far as I can remember the number of columns in the left side matrix must match the number of rows in the right side matrix which is not the case here. The entire formula is here in case it helps put things into context.
Any help would be appreciated.

Best Answer

This should be done on a column vector. The transposed vector on the left will then be a row vector, with the column vector on the right, thus making the product well defined.

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[Math] Transpose a square matrix code

I've just implemented the code in Java (adding the necessary code to feed in the matrix you gave as an example, implement the get and put methods and print the results) and it works fine for me, so there must be a problem elsewhere in your code. I've put my code below. It's not textbook, but it proves the point: there's nothing wrong with your transpose method. Run it and God permitting, you'll see that.

// program which transposes a matrix M and prints M and its transpose

public class Matrix {

    double[][] M = new double[3][3];

    static int mat_size=3;

    public Matrix()
    {
        M[0][0]=1;
        M[0][1]=0;
        M[0][2]=0;
        M[1][0]=5;
        M[1][1]=1;
        M[1][2]=0;
        M[2][0]=6;
        M[2][1]=5;
        M[2][2]=1;

        System.out.println("M");

        printMatrix();

        transpose();

        System.out.println("M'");

        printMatrix();

    }

    public static void main(String[] args) {

        new Matrix();

    }

    public void transpose()
    {
        System.out.println();

        for(int i = 0; i < mat_size; ++i) {
            for(int j = 0; j < i ; ++j) {
                double tmpJI = get(j, i);
                put(j, i, get(i, j));
                put(i, j, tmpJI);
            }
        }
    }



    public void printMatrix()
    {

        System.out.println();

        for(int i = 0; i < mat_size; ++i) {
            for(int j = 0; j < mat_size ; ++j) {
                System.out.print((int)get(i,j)+" ");
            }
            System.out.println();
        }

    }

    private void put(int i, int j, double d) {

        M[i][j]=d;

    }

    private double get(int i, int j) 

        return M[i][j];

    }

}

How would the most general $2 \times 2$ normal matrix look like

You can also characterize a normal matrix $A$ by the fact there exists an unitary matrix such that $U^{-1}AU=D$ where $D$ is diagonal.

Unitary $2x2$ matrices are well characterized, see for instance the wiki page: https://en.wikipedia.org/wiki/Unitary_matrix

$\exists (\alpha,\beta,\xi,\zeta)\in\mathbb R^4$ such that $U=\begin{bmatrix}\cos \alpha & -\sin \alpha\\\sin \alpha & \cos \alpha\end{bmatrix}\begin{bmatrix}e^{i\xi} & 0\\0 & e^{i\zeta}\end{bmatrix}\begin{bmatrix}\cos \beta & -\sin \beta\\\sin \beta & \cos \beta\end{bmatrix}$

$U=R(\alpha).\operatorname{diag}(e^{i\xi},e^{i\zeta}).R(\beta)$

$U^{-1}=R(-\beta).\operatorname{diag}(e^{-i\xi},e^{-i\zeta}).R(-\beta)$

Now you can take $D=\begin{bmatrix}z_1 & 0\\0 & z_2\end{bmatrix}$

Multiplying all these won't probably be very nice looking, but you get to parametrize any normal matrix with $4$ reals and $2$ complexes.

Yet, somehow, this isn't very satisfying for me, the dimension on $\mathbb R$ still appear to be $8$ while C.Haattingh showed that we need $|b|^2=|c|^2$ which can be rewritten $b=re^{it_b}$ and $c=re^{it_c}$ and we reduced the dimension by $1$ already. So I suspect that my $D$ and $(\xi,\zeta)$ are in fact not independent... Is it possible to fix that ?

Best Answer

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[Math] Transpose a square matrix code

How would the most general $2 \times 2$ normal matrix look like

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