Diagonalization Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice diagonalization: recognizing when a matrix has an eigenbasis, reading and building \(A=PDP^{-1}\), matching eigenvectors in \(P\) with eigenvalues in \(D\), using distinct eigenvalues as a fast sufficient test, checking repeated eigenvalues through geometric multiplicity, spotting Jordan-block traps, computing powers as \(A^n=PD^nP^{-1}\), and using eigenvalues for trace, determinant, rank, invertibility, projections, nilpotent cases, and minimal-polynomial checks. If you want a refresher, open the lesson for mentally followable examples and checks.
How this diagonalization practice works
1. Take the quiz: answer eigenbasis, similarity, powers, repeated eigenvalue, and matrix invariant questions at the top of the page.
2. Open the lesson: review what \(A=PDP^{-1}\) means, how to test for enough eigenvectors, and how to use the diagonal form.
3. Retry: return to the quiz and ask whether the matrix has a full basis of eigenvectors.
What you will learn in the diagonalization lesson
Meaning of \(A=PDP^{-1}\)
Diagonalizable: there is a basis made of eigenvectors
\(P\): columns are eigenvectors in the chosen order
\(D\): diagonal entries are the matching eigenvalues
Tests for diagonalizability
In dimension \(n\), diagonalization needs \(n\) linearly independent eigenvectors
Repeated eigenvalues require eigenspace dimensions, not just the characteristic polynomial
Constructing and using the form
Build \(P\) from an eigenbasis and put matching eigenvalues on \(D\)
Use \(A^n=PD^nP^{-1}\) because diagonal powers are entry-by-entry
Trace, determinant, rank, and invertibility become quick diagonal checks
Structure and traps
A nontrivial Jordan block has too few eigenvectors and is not diagonalizable
A diagonalizable matrix with one eigenvalue \(\lambda\) is \(\lambda I\)
The field matters: some real matrices diagonalize only after allowing complex eigenvectors
Back to the quiz
When you are ready, return to the quiz at the top of the page and keep practicing diagonalization.
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Advanced Linear Algebra
Diagonalization Lesson
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Lesson overview
Purpose: Learn to decide whether a matrix is diagonalizable, build the form \(A=PDP^{-1}\) when possible, and use the diagonal form for powers, invariants, and structural conclusions. The central question is always: do the eigenspaces provide a basis of the whole space?
Success criteria
State that diagonalizable means having a basis of eigenvectors.
Explain the roles of \(P\), \(D\), and \(P^{-1}\) in \(A=PDP^{-1}\).
Use distinct eigenvalues as a quick sufficient test.
For repeated eigenvalues, compare algebraic and geometric multiplicity.
Construct \(P\) and \(D\) in matching order.
Compute powers through \(A^n=PD^nP^{-1}\).
Use diagonal entries to read trace, determinant, rank, invertibility, and simple polynomial relations.
Recognize common traps involving Jordan blocks, field choice, and repeated roots.
Key vocabulary
Diagonalizable: similar to a diagonal matrix, equivalently diagonal in some eigenbasis.
Eigenbasis: a basis consisting entirely of eigenvectors of \(A\).
Similarity: \(A=PDP^{-1}\), meaning \(A\) and \(D\) represent the same linear map in different bases.
Algebraic multiplicity: multiplicity of an eigenvalue as a root of the characteristic polynomial.
Geometric multiplicity: dimension of the eigenspace \(E_\lambda\).
Minimal polynomial test: diagonalizable exactly when the minimal polynomial splits with no repeated root.
Quick pre-check
Pre-check 1: A matrix \(A\) is diagonalizable when it has:
Hint: Diagonalization is about changing to a basis where the matrix acts by scaling each basis vector.
Pre-check 2: If \(A=PDP^{-1}\), what are the diagonal entries of \(D\)?
Hint: Each column of \(P\) is scaled by its matching diagonal entry.
An eigenbasis makes the matrix diagonal
Learning goal: Connect the formula \(A=PDP^{-1}\) with a basis of eigenvectors and understand why the order of columns matters.
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 \(P^{-1}\).
Worked example
Example: Diagonalize the swap matrix \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\).
The vectors \((1,1)\) and \((1,-1)\) are eigenvectors: \(A(1,1)=(1,1)\) and \(A(1,-1)=(-1,1)=-(1,-1)\). Thus \(P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\) and \(D=\operatorname{diag}(1,-1)\), with columns matched to diagonal entries. Since the two eigenvectors are independent, \(A=PDP^{-1}\).
Try it
Try it 1: In \(A=PDP^{-1}\), what do the columns of \(P\) contain?
Hint: Multiplying \(A\) by a column of \(P\) should scale that same column.
Try it 2: If the first column of \(P\) has eigenvalue \(2\) and the second has eigenvalue \(5\), what is \(D\)?
Hint: Keep the diagonal entries in the same order as the eigenvectors in \(P\).
Diagonalizable means enough independent eigenvectors
Learning goal: Decide diagonalizability by counting independent eigenvectors, not just by finding eigenvalues.
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 \(n\).
Worked example
Example: A \(3\times3\) matrix has eigenvalues \(1\), \(2\), and \(5\). What follows?
The three eigenvalues are distinct, so there are three independent eigenvectors. In dimension \(3\), that is a full eigenbasis. The matrix is diagonalizable.
Try it
Try it 1: If a \(3\times3\) matrix has three distinct eigenvalues, what follows?
Hint: Eigenvectors for distinct eigenvalues are linearly independent.
Try it 2: A diagonalizable \(4\times4\) matrix must have how many eigenvectors in a basis?
Hint: An eigenbasis of a 4-dimensional space has four basis vectors.
Multiplicity is not enough
Learning goal: Separate algebraic multiplicity from geometric multiplicity and recognize the Jordan-block obstruction.
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.
Worked example
Example: Is \(J=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) diagonalizable?
The only eigenvalue is \(1\). Solve \((J-I)v=0\): \(\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0\), so \(y=0\). The eigenspace is \(\operatorname{span}\{(1,0)\}\), only one-dimensional. A \(2\times2\) matrix needs two independent eigenvectors, so \(J\) is not diagonalizable.
Try it
Try it 1: Is \(\begin{pmatrix}1&1\\0&1\end{pmatrix}\) diagonalizable?
Hint: Compute the eigenspace for the repeated eigenvalue \(1\).
Try it 2: A diagonalizable matrix with all eigenvalues equal to \(5\) is:
Hint: In an eigenbasis the diagonal matrix would be \(5I\), and \(P(5I)P^{-1}=5I\).
Match each eigenvector with its eigenvalue
Learning goal: Construct a diagonalization carefully and avoid mismatching the order of \(P\) and \(D\).
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.
Worked example
Example: For \(A=\begin{pmatrix}2&1\\0&3\end{pmatrix}\), build \(P\) and \(D\).
For \(\lambda=2\), \(A(1,0)=2(1,0)\), so take \(v_1=(1,0)\). For \(\lambda=3\), solve \((A-3I)v=0\), giving \(y=x\), so take \(v_2=(1,1)\). Then \[P=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad D=\operatorname{diag}(2,3).\] The columns are independent, and \(AP=PD\), so \(A=PDP^{-1}\).
Try it
Try it 1: If the first column of \(P\) is an eigenvector with eigenvalue \(4\) and the second column has eigenvalue \(-1\), what is \(D\)?
Hint: The first diagonal entry matches the first column, and the second diagonal entry matches the second column.
Try it 2: If \(A=PDP^{-1}\), \(P\) must be:
Hint: The formula contains \(P^{-1}\).
Diagonalization turns powers into entry-by-entry powers
Learning goal: Use diagonalization to compute powers and read eigenvalue effects quickly.
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.
Worked example
Example: If \(D=\operatorname{diag}(2,-1)\), compute \(D^3\).
Powers of a diagonal matrix are taken entry by entry: \[D^3=\operatorname{diag}(2^3,(-1)^3)=\operatorname{diag}(8,-1).\] Therefore, if \(A=PDP^{-1}\), then \(A^3=P\operatorname{diag}(8,-1)P^{-1}\).
Try it
Try it 1: If \(A=PDP^{-1}\), what is \(A^3\)?
Hint: The middle factors \(P^{-1}P\) cancel.
Try it 2: If \(A\) is diagonalizable with eigenvalues \(2\) and \(3\), what are the eigenvalues of \(A^2\)?
Hint: Squaring \(A\) squares the diagonal entries in a diagonal form.
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
Learning goal: Translate equations involving a diagonalizable matrix into equations for its eigenvalues.
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 \(\lambda^2=\lambda\).
Worked example
Example: Read two facts from diagonal entries: if \(A\) is diagonalizable and \(A^2=A\), what eigenvalues are possible? If another diagonalizable matrix has eigenvalues \(2,3,4\), what are its trace and determinant?
In a diagonal form, \(D^2=D\). So every diagonal entry \(\lambda\) satisfies \(\lambda^2=\lambda\), or \(\lambda(\lambda-1)=0\). The only possible eigenvalues are \(0\) and \(1\). For eigenvalues \(2,3,4\), the trace is \(2+3+4=9\) and the determinant is \(2\cdot3\cdot4=24\).
Try it
Try it 1: If \(A\) is diagonalizable and \(A^2=0\), then \(A\) is:
Hint: Every eigenvalue \(\lambda\) satisfies \(\lambda^2=0\), so \(D\) must be the zero diagonal matrix.
Try it 2: If \(D=\operatorname{diag}(0,1,2)\), what is the rank of \(D\)?
Hint: Count the nonzero diagonal entries.
Splitting and repeated roots matter
Learning goal: Understand field dependence, minimal-polynomial tests, and finish with a reliable final checklist.
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.
Worked example
Example: Compare the rotation \(R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) over \(\mathbb{R}\) and over \(\mathbb{C}\).
The characteristic equation is \(\lambda^2+1=0\). Over \(\mathbb{R}\), it has no roots, so there is no real eigenbasis and no real diagonalization. Over \(\mathbb{C}\), the eigenvalues are \(i\) and \(-i\), distinct, so the matrix is diagonalizable over \(\mathbb{C}\).
Try it
Try it 1: If a matrix is diagonalizable, its minimal polynomial has:
Hint: Diagonalization rules out repeated Jordan-block behavior.
Try it 2: Is a nontrivial \(2\times2\) Jordan block diagonalizable?
Hint: It has only one eigenvector direction for a repeated eigenvalue.
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.
Trace, determinant, rank, invertibility, and polynomial equations can be read from the diagonal form.
The field and the minimal polynomial matter.
Next step: Close this lesson and try the quiz again. For each question, first decide whether it asks for an eigenbasis test, a construction of \(P\) and \(D\), a repeated-eigenvalue trap, or a consequence of diagonal entries.
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