Distinguish diagonalizability from merely having characteristic roots.
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Try
If A=PDP⁻¹ with diagonal D, repeated multiplication simplifies: A²=PD²P⁻¹ because the inner P⁻¹P cancels. In an eigenvector basis, each coordinate evolves independently by a scalar factor.
Predict
Predict what P must contain. Its columns must be a basis of eigenvectors; the matching eigenvalues appear on the diagonal of D in the same order. Having n characteristic roots counted with multiplicity does not guarantee n independent eigenvectors.
Experiment
Use A=[[2,1],[0,3]], P=[[1,1],[0,1]], D=diag(2,3). Calculate AP and PD and compare. Compute A² both directly and through PD²P⁻¹. The agreement checks the representation; the eigenbasis theorem explains why it works for all powers.
Understand
Worked failure: J=[[1,1],[0,1]] has characteristic polynomial (λ−1)², but (J−I)(x,y)=0 forces y=0. There is only one independent eigenvector direction, so J is not diagonalizable. In contrast the identity also has a repeated eigenvalue but every vector is in its eigenspace, giving a full basis.
The criterion is exact: A is diagonalizable over R if and only if it has a real eigenbasis. If it does, AP=PD follows column by column; multiplying by P⁻¹ gives the formula. Conversely that formula makes P’s columns eigenvectors. With n distinct real eigenvalues, their independent eigenvectors suffice.
Invalid conclusion: a failed diagonalization means powers cannot be computed. For J=I+N with N²=0, the binomial expansion gives Jᵏ=I+kN.
Apply
Track basis order and eigenvalue order together. For a negative eigenvalue, odd and even powers have different signs. To disprove diagonalizability, determine a deficient eigenspace dimension rather than merely observing that the original matrix is not diagonal.
Continue
Depth: symmetric real matrices always admit an orthonormal eigenbasis, a spectral theorem developed fully in Linear Algebra II. General canonical forms handle defective matrices. This lesson introduces the obstruction without claiming coverage of Jordan theory.
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Practice and reasoning
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A=diag(2,−1). Enter the first coordinate of A³(3,4).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For J=[[1,1],[0,1]], enter the (1,2) entry of J⁴.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
An n×n real matrix is diagonalizable over R exactly when:
A repeated eigenvalue:
Diagonalize A=[[2,1],[0,3]] using P with columns (1,0),(1,1), and verify AP=PD.
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Prove J=[[1,1],[0,1]] is not diagonalizable over R.
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