An eigenbasis makes the matrix diagonal

Key idea

Suppose \(v_1,\dots,v_n\) is a basis and \(Av_i=\lambda_i v_i\). Put these vectors into the columns of \(P\), and put the matching eigenvalues on the diagonal of \(D\): \[P=\begin{pmatrix}|& &|\\ v_1&\cdots&v_n\\ |& &|\end{pmatrix},\qquad D=\operatorname{diag}(\lambda_1,\dots,\lambda_n).\] Then \(AP=PD\), so \(A=PDP^{-1}\). The formula is not magic: it says that in the eigenbasis, \(A\) only scales coordinates.

Recognition checklist

• Find eigenvalues and eigenspaces.
• Choose enough independent eigenvectors to form a basis.
• Place the eigenvectors as columns of \(P\).
• Place the matching eigenvalues in the same order on \(D\).
• Check \(AP=PD\) before computing \(P^{-1}\).

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Diagonalizable means enough independent eigenvectors

Key idea

For an \(n\times n\) matrix, diagonalization requires \(n\) linearly independent eigenvectors. Distinct eigenvalues are helpful because eigenvectors belonging to distinct eigenvalues are automatically linearly independent. Therefore, \(n\) distinct eigenvalues in dimension \(n\) guarantee diagonalizability. Repeated eigenvalues are not automatically bad, but they require checking eigenspace dimensions.

Recognition checklist

• If the matrix has \(n\) distinct eigenvalues over the field you are using, it is diagonalizable.
• If eigenvalues repeat, compute each eigenspace \(E_\lambda=\ker(A-\lambda I)\).
• Add the dimensions of the eigenspaces.
• The matrix is diagonalizable exactly when that sum is \(n\).

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Multiplicity is not enough

Key idea

A repeated eigenvalue can still be diagonalizable, but only if its eigenspace is large enough. For each eigenvalue \(\lambda\), \[\dim E_\lambda\le \text{algebraic multiplicity of }\lambda.\] Diagonalization needs equality for every eigenvalue, so the geometric multiplicities add to \(n\). A nontrivial Jordan block fails because it has only one eigenvector direction for a repeated eigenvalue.

Common traps

• A repeated root of the characteristic polynomial does not automatically mean not diagonalizable.
• A triangular matrix is not automatically diagonalizable.
• A matrix with one eigenvalue can be diagonalizable only if it is already a scalar matrix on the whole space.
• A nontrivial Jordan block has too few eigenvectors.

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Match each eigenvector with its eigenvalue

Key idea

After finding an eigenbasis, diagonalization is mostly bookkeeping. If the columns of \(P\) are \(v_1,\dots,v_n\), then the diagonal entries of \(D\) must be \(\lambda_1,\dots,\lambda_n\), where \(Av_i=\lambda_i v_i\). Changing the order of columns is allowed, but \(D\) must change in the same order.

Recognition checklist

• Choose independent eigenvectors until you have a basis.
• Write them as columns of \(P\).
• Write matching eigenvalues on the diagonal of \(D\).
• Check \(AP=PD\), which is often faster than multiplying \(PDP^{-1}\).
• Only compute \(P^{-1}\) when a later calculation actually needs it.

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Diagonalization turns powers into entry-by-entry powers

Key idea

If \(A=PDP^{-1}\), then repeated multiplication cancels the middle factors: \[A^n=(PDP^{-1})^n=PD^nP^{-1}.\] For \(D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\), \[D^n=\operatorname{diag}(\lambda_1^n,\dots,\lambda_n^n).\] The same idea works for many polynomials in \(A\): apply the polynomial to each diagonal entry.

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Summary

• \(A^n=PD^nP^{-1}\), not \(P^nD^nP^{-n}\).
• Diagonal powers are entry-by-entry.
• If no eigenvalue is \(0\), then \(A^{-1}=PD^{-1}P^{-1}\).
• For example, \(\operatorname{diag}(1,2)^{-1}=\operatorname{diag}(1,1/2)\).

Use diagonal entries to read structure

Key idea

When \(A=PDP^{-1}\), many structural facts reduce to the diagonal entries. \(A\) is invertible exactly when every eigenvalue is nonzero. The trace is the sum of eigenvalues, the determinant is their product, and the rank is the number of nonzero diagonal entries in \(D\). If \(A\) satisfies a polynomial equation such as \(A^2=A\), then each eigenvalue satisfies the same scalar equation \(\lambda^2=\lambda\).

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Splitting and repeated roots matter

Key idea

Diagonalization depends on the field. Over \(\mathbb{R}\), a matrix with nonreal eigenvalues cannot have a real eigenbasis. Over \(\mathbb{C}\), those eigenvalues may become available. A compact high-level test is: a matrix is diagonalizable over a field exactly when its minimal polynomial splits over that field and has no repeated root.

Common traps

• Similar matrices share trace, determinant, characteristic polynomial, and eigenvalues, but not the same entries.
• Diagonalizable does not mean diagonal in the original basis.
• Having eigenvalues is not enough; the eigenspaces must supply a basis.
• Distinct eigenvalues are sufficient, not necessary.
• The minimal polynomial must split into distinct linear factors over the chosen field.

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Final recap

• Diagonalizable means there is an eigenbasis.
• \(A=PDP^{-1}\) with columns of \(P\) equal to eigenvectors and diagonal entries of \(D\) equal to matching eigenvalues.
• \(n\) distinct eigenvalues in dimension \(n\) guarantee diagonalizability.
• Repeated eigenvalues require eigenspace dimension checks.
• Nontrivial Jordan blocks are not diagonalizable.
• Powers satisfy \(A^n=PD^nP^{-1}\).
• Trace, determinant, rank, invertibility, and polynomial equations can be read from the diagonal form.
• The field and the minimal polynomial matter.