Can this symmetric matrix be an orthogonal matrix?

Question

So the question is to prove whether this matrix M is orthogonal?

$$ M = \begin{bmatrix} a & k & k \\ k & a & k \\ k & k & a \\ \end{bmatrix} $$

My attempt to find the inverse of M :

$$det(M)=(a+2k)(a-k)(a-k)$$

The inverse of $M$ is :

$$ \begin{matrix} (a+k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k))\\ (-k)/((a-k)(a+2k)) & (a+k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k))\\ (-k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k)) & (a+k)/((a-k)(a+2k))\\ \end{matrix} $$

$M$ is orthogonal if $M^T = M^{-1}$

So I end up with a system of 2 equations to solve :

$a=\dfrac{a+k}{(a-k)(a+2k)}$ and $k=\dfrac{-k}{(a-k)(a+2k)}$

and now I am stucked.

I would really appreciate feedbacks about whether my steps are all correct and if yes, what to do next ?

Answer 1

Since $M$ is symmetric, $M=M^T$, so to check orthogonality, we compute $MM^T=M^2$ and get $$ \begin{bmatrix} a^2+2k^2&2ak+k^2&2ak+k^2\\ 2ak+k^2&a^2+2k^2&2ak+k^2\\ 2ak+k^2&2ak+k^2&a^2+2k^2 \end{bmatrix} $$

Therefore, you need $2ak+k^2=0$ and $a^2+2k^2=1$. Factoring the first equation, we have $k(2a+k)=0$. Therefore, either $k=0$ or $k=-2a$.

• When $k=0$, the second equation simplifies to $a^2=1$, so $a=\pm 1$ gives orthogonality.

• When $k=-2a$, the second equation simplifies to $9a^2=1$, so $a=\pm\frac{1}{3}$ with $k=\mp \frac{2}{3}$ gives orthogonality.

Answer 2

So the question is to prove whether this matrix M is orthogonal?

It's not clear what exactly what you mean by this. Perhaps your question is:

Is $M$ orthogonal for all $a,k \in \Bbb R$?

The answer to this is clearly no. For instance: if $a = k$, then $M$ fails to be invertible, let alone orthogonal. It is also notable that orthogonal matrices have a determinant of $\pm 1$.

Perhaps your question is:

For which $a,k \in \Bbb R$ is $M$ orthogonal?

Note that $M$ is orthogonal if and only if $MM^T = I$. However, $M$ is symmetric, so $M = M^T$, and the above can be rewritten as $M^2 = I$. To that end: $$ M^2 = \pmatrix{a^2 + 2k^2 & 2ak + k^2 & 2ak + k^2 \\ \vdots & \ddots} $$ So, our matrix will be orthogonal if and only if $M$ is the identity matrix, which is to say that $$ a^2 + 2k^2 = 1\\ 2ak + k^2 = 0 $$ The last equation can be written as $k(2a + k) = 0$. If $k = 0$, then the first equation becomes $a^2 = 1$ so that $a = \pm 1$ are the values for which $M$ is orthogonal. If $k = -2a$, then $9a^2 = 1 \implies a = \pm 1/3$. So, we have the additional solutions $a=1/3,k=-2/3$ and $a = -1/3, k = 2/3$.

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This is all made much easier using eigenvalues, and noting that $M$ has the form $$ M = k xx^T + (a - k)I $$ where $x$ is the column vector of $1$s and $I$ is the identity matrix.

Answer 3

It would be easier to solve $$A^\top A = I $$.

$$ \begin{eqnarray*} \left[\begin{array}{ccc} a & k & k\\ k & a & k\\ k & k & a \end{array}\right]\left[\begin{array}{ccc} a & k & k\\ k & a & k\\ k & k & a \end{array}\right] & = & \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\\ \left[\begin{array}{ccc} a^{2}+2k^{2} & 2ak+k^{2} & 2ak+k^{2}\\ 2ak+k^{2} & a^{2}+2k^{2} & 2ak+k^{2}\\ 2ak+k^{2} & 2ak+k^{2} & a^{2}+2k^{2} \end{array}\right] & = & \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right] \end{eqnarray*} $$ Now you can solve the following equations. $$ \begin{eqnarray*} a^{2}+2k^{2} & = & 1\\ 2ak+k^{2} & = & 0 \end{eqnarray*} $$