How can I say that this matrix is invertible?

Question

Let $A_{ij}=1$ if $i=j$ and let $A_{ij}=\frac{1}{2}$ if $i \neq j$. How can I formally argue that $A$ is an invertible matrix? I can made for particular examples, but I don't know how to argue for the general case.

Answer 1

$$ \begin{bmatrix} 1&\frac12&\frac12&\cdots&\frac12\\ \frac12&1&\frac12&\cdots&\frac12\\ \frac12&\frac12&1&\cdots&\frac12\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \frac12&\frac12&\frac12&\cdots&1\\ \end{bmatrix} $$ subtract $\frac1n\times$ the sum of the $n-1$ right columns from column $1$ and we get $$ \begin{bmatrix} \frac{n+1}{2n}&\frac12&\frac12&\cdots&\frac12\\ 0&1&\frac12&\cdots&\frac12\\ 0&\frac12&1&\cdots&\frac12\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&\frac12&\frac12&\cdots&1\\ \end{bmatrix} $$ subtract $\frac1{n-1}\times$ the sum of the $n-2$ right columns from column $2$ and we get $$ \begin{bmatrix} \frac{n+1}{2n}&\frac1{2(n-1)}&\frac12&\cdots&\frac12\\ 0&\frac{n}{2(n-1)}&\frac12&\cdots&\frac12\\ 0&0&1&\cdots&\frac12\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0 &\frac12&\cdots&1\\ \end{bmatrix} $$ Continuing for the rest of the columns, we end up with $$ \begin{bmatrix} \frac{n+1}{2n}&\frac1{2(n-1)}&\frac1{2(n-2)}&\cdots&\frac12\\ 0&\frac{n}{2(n-1)}&\frac1{2(n-2)}&\cdots&\frac12\\ 0&0&\frac{n-1}{2(n-2)}&\cdots&\frac12\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1\\ \end{bmatrix} $$ Computing the determinant of the upper triangular matrix gives $$ \frac{n+1}{2^n} $$

Answer 2

Here is a short way to compute its determinant:

Let $J$ be the matrix of dimension $n$ with all coefficients equal to $\frac12$. As this matrix has rank $1$, its kernel has dimension $n-1$, so, using Vieta's relations, we see that its characteristic polynomial is $$\det (XI_n-J)=X^n -\operatorname{Tr}(J) X^{n-1}=X^{n-1}\Bigl(X-\frac n2\Bigr).$$ Now consider the matrix $A(x)$ which is the same as your matrix, except the $1$s on the diagonal are replaced with a variable $x$. We have \begin{align} \det A(x)&=\det\bigl(J+\bigl(x-\tfrac12\bigr) I_n\bigr)=(-1)^n\det\bigl(\bigl(\tfrac12-x)I_n-J\bigr)\bigr) \\ &=(-1)^n\bigl(\tfrac12-x\bigr)^{n-1}\bigl(\tfrac12-x-\tfrac n2\bigr)=\bigl(x-\tfrac12\bigr)^{n-1}\bigl(x+\tfrac{n-1}2\bigr). \end{align} What we want is $$A(1)=\frac1{2^{n-1}}\frac{n+1}2=\frac{n+1}{2^n}.$$

Answer 3

Notice that for every $x = (x_1, \ldots, x_n) \in \mathbb{R}^n, x \ne 0$ we have $$\langle Ax,x\rangle = \sum_{i=1}^n x_i^2 + \sum_{i\ne j}x_ix_j = \frac12\left(\sum_{i=1}^nx_i\right)^2 + \frac12\sum_{i=1}^nx_i^2 > 0$$

so $A$ is positive definite. In particular, $A$ is invertible.

Answer 4

Note that $$2A\:=\:\mathbb 1_n+\:\begin{pmatrix}1\\\vdots\\1\end{pmatrix}(1,\cdots,1)$$ where $\mathbb 1_n$ is the identity matrix. The eigenvalues to the $n$ eigenvectors $$\begin{pmatrix}1\\1\\1\\\vdots\\1\end{pmatrix},\; \begin{pmatrix}-1\\1\\0\\\vdots\\0\end{pmatrix},\; \begin{pmatrix}0\\-1\\1\\0\\\vdots\end{pmatrix},\; \dots, \begin{pmatrix}0\\\vdots\\0\\-1\\1\end{pmatrix}$$ are non-zero.