Suppose a 7×7 matrix A has three distinct eigenvalues, support further that one of the eigenspaces is three dimensional, and one of the eigenspaces is two dimensional. Determine if A must be diagonalizable, is never diagonalizable, or could be diagonalizable under certain additional conditions. Explain your answer (Including any additional conditions you might require.)
Conceptual problems are not my strong suit. I know that the diagonal matrix must be 7×7 and the diagonal entries are the eigenvalues which there are three of. Where do i go with this information?
Best Answer
Diagonalizability means there is a basis for your space for which the matrix acts on each basis vector as scalar multiplication. In other words, a basis of eigenvectors. Therefore a matrix is diagonalizable if and only if it has as many independent eigenvectors as the dimension of the space. Looking separately at each eigenvalue, we can say a matrix is diagonalizable if and only if for each eigenvalue the geometric multiplicity (dimension of eigenspace) matches the algebraic multiplicity (number of times it is a root of the characteristic polynomial).
If it's a 7x7 matrix; the characteristic polynomial will have degree 7. So by the fundamental theorem of algebra, the algebraic multiplicities sum to 7 if we work over the complexes. You have two eigenvalues with geometric multiplicities are 2 and 3. If the geometric multiplicity of the third eigenvalue is 2 (i.e. if the third eigenvalue has 2 linearly independent eigenvectors) then the total number of independent eigenvectors is 7, matching the total dimension of the space, so the matrix is diagonalizable. If not, if instead the third eigenvalue has a only a one-dimensional eigenspace (this is the only other possibility; it cannot have eigenspace of dimension 0 or dimension greater than 2, do you see why?) then the matrix is not diagonalizable, since then you have $3+2+1=6$ independent eigenvectors, not enough to span your 7-dimensional space. You would determine the number of independent eigenvectors by computing the nullspace of $A-\lambda_3$. And counting its dimension.