4.5.31 · D5Linear Algebra (Full)
Question bank — Diagonalization — conditions, procedure
Recall the vocabulary you must never blur:
- Algebraic multiplicity — how many times is a root of .
- Geometric multiplicity — how many independent eigenvectors actually supplies.
- Diagonalizable means with invertible and diagonal, which forces to own independent eigenvectors.
True or false — justify
A matrix with a repeated eigenvalue is never diagonalizable
False — repeats yet gives two independent eigenvectors (); repetition only threatens diagonalizability when .
Every matrix with distinct eigenvalues is diagonalizable
True — distinct eigenvalues force their eigenvectors to be linearly independent, handing you independent columns for .
If a matrix is diagonalizable then it must have distinct eigenvalues
False — the converse fails; is diagonalizable with a single repeated eigenvalue, so "distinct" is sufficient but not necessary.
A real matrix always has a real diagonalization
False — the rotation has eigenvalues , so it diagonalizes only over , not .
Diagonalizable matrices are exactly the invertible matrices
False — the two notions are independent; is diagonal (hence diagonalizable) but not invertible, and a shear is invertible but not diagonalizable.
If is diagonalizable then so is
True — if then with the same , and is diagonal (see Matrix Powers and Exponentials).
Two similar matrices always have the same eigenvalues
True — Similar Matrices share the same Characteristic Polynomial, so their roots (the eigenvalues) coincide, though their eigenvectors need not.
Every symmetric real matrix is diagonalizable
True — the Spectral Theorem guarantees a real orthonormal eigenbasis, so for every eigenvalue automatically.
Scaling an eigenvector changes the diagonal matrix
False — an eigenvector and any nonzero multiple of it share the same eigenvalue, so only changes while (and the pairing) stays fixed.
A matrix and its transpose are diagonalizable together or not at all
True — and share the same characteristic polynomial and the same pattern, so one is diagonalizable exactly when the other is.
If every eigenvalue of equals then is the zero matrix
False — a nilpotent shear like has only yet is nonzero (and defective, so not diagonalizable).
Spot the error
" has characteristic polynomial , so it must be defective."
Error — algebraic multiplicity says nothing alone; you must compute ; if (e.g. ) it is perfectly diagonalizable.
"I put eigenvalues in in increasing order and stacked the eigenvectors in however I liked."
Error — column of must be an eigenvector for the eigenvalue in ; scrambling the pairing silently breaks even though the same eigen-data appears.
" can exceed when an eigenvalue is very repeated."
Error — the inequality is always ; geometric multiplicity can never outrun algebraic multiplicity.
"Since contains eigenvectors, any set of eigenvectors makes a valid ."
Error — the eigenvectors must be linearly independent so that is invertible; a defective matrix's eigenvectors don't span, so no valid exists.
" means equals , just written differently."
Error — and are only similar (same transformation in different bases via Change of Basis); they are equal only when , i.e. when was already diagonal.
"The characteristic polynomial has degree , so there are always real eigenvalues, hence eigenvectors."
Error — the roots may be complex or repeated, and repeated roots may under-supply eigenvectors, so neither "real" nor " independent eigenvectors" is guaranteed.
Why questions
Why must the columns of be eigenvectors and not any convenient basis?
Because expanding column by column gives , which is exactly the eigenvalue equation — diagonalization is forced, not chosen.
Why does the condition require independent eigenvectors rather than just eigenvectors?
Because must be invertible, and an matrix is invertible if and only if its columns are linearly independent; dependent columns give a singular and no .
Why do distinct eigenvalues automatically give independence?
Eigenvectors for different eigenvalues live in different stretch-directions and can't be built from one another, so any combination forcing a dependency collapses to the zero vector — they must be independent.
Why is diagonalization so useful for computing ?
In the eigenbasis only stretches each axis, so and is just each diagonal entry to the th power — a hard repeated multiply becomes one line (see Matrix Powers and Exponentials).
Why does a pure shear fail to be diagonalizable?
A shear fixes one direction and twists everything else, so it owns only a single stretch-direction; with just one independent eigenvector for a map there's no full eigenbasis to redraw axes along.
Why can a real matrix be diagonalizable over but not over ?
Rotations have no real fixed direction, so their eigenvalues are complex (); the eigenvectors are complex too, meaning the diagonalizing basis simply doesn't exist inside .
Why doesn't the order you list eigenvectors change whether is diagonalizable?
Reordering columns of (and matching ) just relabels the basis; the set is still independent and still spans, so diagonalizability is a property of the set, not its order.
Edge cases
Is the zero matrix diagonalizable?
Yes — it is already diagonal, with of algebraic multiplicity and giving .
Is a matrix always diagonalizable?
Yes — a single number is trivially diagonal, its lone eigenvector is any nonzero scalar, and never fails.
If has a zero eigenvalue, can it still be diagonalizable?
Yes — is a legitimate stretch factor (it flattens that direction); e.g. a projection like is diagonal with eigenvalues and .
Is a diagonal matrix with a repeated entry, like , diagonalizable?
Yes — it is already diagonal; the repeated eigenvalue has a full -dimensional eigenspace (), so nothing is defective.
Does a matrix with for some eigenvalues but not all still diagonalize?
No — the theorem demands for every eigenvalue; a single deficient eigenvalue costs you an independent eigenvector and breaks the count for .
When has a full set of eigenvalues but one eigenspace is deficient, what structure describes it instead?
It is not diagonalizable but is still similar to a nearly-diagonal form with s just above the diagonal — the Jordan Normal Form, the canonical home for defective matrices.
Recall One-line survival test
diagonalizes for every eigenvalue, the number of independent eigenvectors it hands you () equals how often it appears as a root (). Distinct eigenvalues and symmetric real matrices pass this automatically; shears and rotations-over- are the classic failures.