Simple way to find determinant of a given matrix

Question

Consider the $(m-1)\times (m-1)$ matrix $M$ shown below. $$ M= \left[ {\begin{array}{cc} 1 & 1/2 & 1/2 & \ldots & 1/2\\ 1/2 & 1 & 1/2 & \ldots & 1/2\\ \vdots \\ 1/2 & 1/2 & 1/2 & \dots & 1 \end{array} } \right]$$

I would like to show that $\det (M)=\frac{m}{2^{m-1}}$. I have tried induction, but the computation gets rather messy. When $m=2$, $\det(M)=1=2/2$, and when $m=2$, $\det(M)=1-1/2^2=3/2^2$.

I know some people are very good at seeing tricks to compute the determinant of a matrix like this without induction. Any ideas?

Thanks!

Add all the rows to the last one. You can take out the common factor and you'll get a row with all 1/2. You can substract it from every other row to get a diagonal matrix — mlainz, Apr 03 '17 at 22:04

Misha Lavrov · Accepted Answer · 2017-04-03T22:09:52.620

3

The determinant is the product of the eigenvalues. We know that $\frac12$ is an eigenvalue with multiplicity at least $m-2$, because $M - \frac12I$ has rank $1$. In addition, the sum of the eigenvalues is the trace of $M$, which is $m-1$. So if $x$ is the remaining eigenvalue, then $x + \frac12(m-2) = m-1$, which means that $x = \frac m2$.

Taking the product of the eigenvalues, we get $\frac m2 \cdot \left(\frac12\right)^{m-2} = \frac{m}{2^{m-1}}$.

Also, Sylvester's determinant identity applies to $2$ times this matrix. If $\mathbf u$ is the vector of all $1$'s, then $2M = I + \mathbf{u}\mathbf{u}^T$, so $\det(2M) = \det(I + \mathbf{u}\mathbf{u}^T) = 1 + \mathbf{u}^T\mathbf{u} = m$. Therefore $\det(M) = \frac1{2^{m-1}} \det(2M) = \frac{m}{2^{m-1}}$.

edited Apr 03 '17 at 22:09

answered Apr 03 '17 at 22:06

Misha Lavrov

142,276

3

Or - by inspection, $(1, 1, \ldots, 1)$ is an eigenvector of $M$ with eigenvalue $\frac{1}{2}(m-2) + 1 = \frac{m}{2}$. – Daniel Schepler Apr 03 '17 at 22:08
@MishaLavrov Thank you! Just a quick question...why does $M-\frac{1}{2}I$ having rank 1 imply that $1/2$ is an eigenvalue of $M$ with multiplicity at least $m-2$? Are you using the Cayley-Hamilton Theorem? Sorry, my linear algebra is a bit rusty... – Sarah Apr 03 '17 at 22:27
2

In linear algebra, the geometric multiplicity--which is the dimension of the eigenspace--of an eigenvalue a is the nullspace of $(M-aI)$. So since the matrix in question is rank 1 which is $m-2$ smaller than full rank, the nullspace is $(m-2)$ dimensional. Then, the algebraic multiplicity of $a$, which is the exponent of $a$ in det(M) or the exponent of (a - lambda) in det(M-lambda I), is always at least the geometric multiplicity. – Elizabeth S. Q. Goodman Apr 03 '17 at 22:30
For matrices of this form I always take the difference of diagonal to off-diagonal for the main eigenvalue, as Misha did here, and then I use [1,1,...1] as an eigenvector to find the remaining eigenvalue (i.e. the sum of entries in any row), as Daniel did. – Elizabeth S. Q. Goodman Apr 03 '17 at 22:31

score 0 · Answer 2 · answered Apr 03 '17 at 22:08

More generally, consider the $n\times n$ matrix $$ M(a)=\begin{bmatrix} a & a & a & \dots & a \\ a & a & a & \dots & a \\ \vdots & \vdots & \ddots & \dots & \vdots \\ a & a & a & \dots & a \end{bmatrix} $$ The characteristic polynomial of $M(a)$ is $$ \det(M(a)-XI)=(0-X)^{n-1}(na-X) $$ Can you see why?

What value can you give to $X$ in order to have your given matrix?

Simple way to find determinant of a given matrix

2 Answers2