I know real symmetric matrices have real eigenvalues, and are orthogonally diagonalizable. But not all are invertible, e.g. a really trivial example:

[0 0]
[0 0]

Is clearly not invertible because the 2 columns are not linearly independent, it has 0's as its eigenvalues, etc. The above example falls into the category where every element is nonnegative, i.e. on [0,infinty).

The identity matrix is also on [0,inf) but is invertible and has nonzero diagonal entries.

As another example,

[1 1]
[1 1]

is not invertible.

So what are necessary/sufficient conditions for a real symmetric matrix to be invertible?

  • 1
    $\begingroup$ Here's a simple matrix with positive entries: $\pmatrix{1&1\\1&1}$. $\endgroup$ Commented Sep 10, 2017 at 17:01
  • $\begingroup$ of course, ok updated the question $\endgroup$
    – dektorpan
    Commented Sep 10, 2017 at 22:48
  • $\begingroup$ When it comes to invertibility, I don't think there's anything special about real symmetric matrices – they are no easier to check for invertibility than are any other kind of matrix. $\endgroup$ Commented Sep 11, 2017 at 3:07
  • $\begingroup$ Are you happy with barman's answer, dek? Any points need clarification? $\endgroup$ Commented Sep 12, 2017 at 7:22

1 Answer 1


Yes, a matrix is invertible if and only if its determinant is not zero. You may have heard of the general linear group $GL(\mathbb{K},n)$ where $\mathbb{K}$ is some field and $n$ is the dimension of the vector space. It denotes the group of invertible matrices.

To see why this determinant criterion works there are several ways. I'm going to write the easiest one out of lazyness. If $A$ is invertible, then exists $B = A^{-1}$ such that $AB =Id$, now $\det(AB) = \det(A)\det(B)=\det(Id) = 1$ so both $\det(A)$ and $\det(B)$ ought to be $\neq 0$.

Also note that the matrices whose determinant is different from zero as you said are injective transformations of vector spaces. Since the determinant represents how the transformation changes the unit area of the space, and an injective linear transofrmation never reduces the space to a lower dimension, the determinant can't be zero.


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