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Wikipedia gives me,

In linear algebra, a square matrix $A$ is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix $P$ such that $P^{−1}AP$ is a diagonal matrix.

However, this condition is not very useful when checking for diagonalizability of a square matrix. Can you suggest me ways to check such a condition when just the matrix is given?

Pritt Balagopal
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  • It's diagonalizable (over, say, the complex numbers – you do have to specify the field!) if there is a basis for ${\bf C}^n$ consisting entirely of eigenvectors of $A$. – Gerry Myerson Nov 12 '17 at 10:12

2 Answers2

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  • If the matrix is symmetric, it is diagonalizable and the obtained matrix $P$ can be chosen to be orthogonal.

  • If all the eigenvalues are distinct, then it is diagonalizable, i.e. An $n\times n$ matrix $A$ is diagonalizable over the field $F$ if its charactheristic polynomial has $n$ distinct roots in $F$.

  • If the sum of the dimension of the eigenspace is equal to $n$, then it is diagonalizable.

Siong Thye Goh
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A square matrix is diagonalizable if and only if its minimal polynomial is squarefree.

This needs a bit of explanation. The base field, the smallest field containing the entries of the matrix, may not contain all the roots of the minimal polynomial; for example, a real matrix and minimal $x^2 + 1.$ Furthermore, enlarging the field does not change the minimal polynomial. However, it may allow us to factor, as in $x^2 + 1 = (x-i)(x+i).$ Here we can see clearly that the two linear factors are distinct. Put another way, the roots of the minimal polynomial in any field extension are distinct.

By Cayley-Hamilton, we do not need to find roots to confirm the minimal polynomial, just find the lowest degree polynomial dividing the characteristic polynomial that annihilates the matrix.

See Proposition 2.8 and Examples 2.9 in Michael Stoll, on pages 5 and 6.

Will Jagy
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