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Linear Algebra – Eigenvalues and Eigenvectors - Document preview page 1

Linear Algebra – Eigenvalues and Eigenvectors - Page 1

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Linear Algebra – Eigenvalues and Eigenvectors

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Linear Algebra – Eigenvalues and Eigenvectors - Page 1 preview imageStudy GuideLinear AlgebraEigenvalues and Eigenvectors1.Determining the Eigenvalues of a Matrix1. Why Do We Care About Eigenvalues?In linear algebra, every linear operator can be represented by multiplying by a square matrix.So instead of thinking about abstract operators, we can just studysquare matrices. That means:Finding the eigenvalues and eigenvectors of a linear operator is the same as finding them for asquare matrix.Throughout this chapter, we assume all matrices aresquare matrices.2. What Is an Eigenvalue?Let’s start with a square matrix (A).A number (λ) (lambda) is called aneigenvalueof (A) if there exists anonzero vector(x) such that:This means:When matrix (A) acts on vector (x),It onlystretches or shrinks the vector,It doesnotchange its direction.That special vector (x) is called aneigenvector.
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Linear Algebra – Eigenvalues and Eigenvectors - Page 2 preview imageStudy Guide3. Turning the Equation Into a Useful FormWe start with:Now rewrite it step by step:Factor out (x):Here:(I) is theidentity matrix.(AλI) means subtract (λ) from every diagonal entry of (A).4. When Does a Nonzero Solution Exist?We want anonzero vector(x).For the systemto have a nonzero solution, the determinant must be zero:This equation is called thecharacteristic equation.The expression:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 3 preview imageStudy Guideis called thecharacteristic polynomial.Important facts:It is a polynomial in (λ).It has degree (n) if (A) is an (n×n) matrix.The roots of this polynomial are theeigenvalues.Example 1:Finding Eigenvalues of a 2×2 MatrixWe are given:Step 1: Form (AλI)Subtract (λ) from each diagonal entry:Step 2: Take the DeterminantNow simplify:This is thecharacteristic polynomial.Step 3:Solve the Characteristic Equation
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Linear Algebra – Eigenvalues and Eigenvectors - Page 4 preview imageStudy GuideSet it equal to zero:Factor:So the eigenvalues are:A Small Note About NotationSometimes books write:instead ofFor:Even-sized matrices → they are exactly thesame.Odd-sized matrices → they differ by a negative sign.But thisdoes not affect the eigenvalues, because both expressions become zero at the samevalues of (λ).Example 2:3×3 Checkerboard MatrixWe are given:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 5 preview imageStudy GuideStep 1: Form (CλI)Subtract (λ) from the diagonal:Step 2:Compute the DeterminantTo simplify the determinant:Add the second row to the third row.Then expand along the first column (Laplace expansion).After simplifying carefully, we get:Step 3:Solve the Characteristic EquationSet equal to zero:This gives:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 6 preview imageStudy GuideSo the eigenvalues are:(λ) (with multiplicity 2)(λ= 3)2.Determining the Eigenvectors of a Matrix1. Big Idea: How Do We Find Eigenvectors?Before finding eigenvectors, we must first know theeigenvalues.Why?Because eigenvectors are found by plugging each eigenvalue into the equation:This can also be written as:Now we solve this system forx.Important:We are looking fornonzero solutions.Those nonzero solutions form theeigenvectors corresponding to that eigenvalue.We repeat the process for each eigenvalue.Example 1:Finding Eigenvectors of a 2×2 MatrixWe use the matrix:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 7 preview imageStudy GuideFrom the previous section, we already found the eigenvalues:Now let’s find the eigenvectors for each one.Eigenvectors forλ = −1We substitute λ = −1 into:So we solve:This gives a system of equations. After simplifying, we obtain:Notice somethingimportant:These two equations arenot independent.They are actually the same equation written twice.That means we have infinitely many solutions.From:So every solution looks like:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 8 preview imageStudy Guidewhich can be written as:wheretis any nonzero scalar.Therefore, the eigenvectors corresponding to λ = −1 are all multiples of:Eigenvectors forλ = −2Now substitute λ = −2 into:After simplifying, we get the system:Again, the equations are not independent.Solving:Thisgives vectors of the form:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 9 preview imageStudy Guidewheret ≠ 0.Therefore, the eigenvectors corresponding to λ = −2 are all multiples of:Understanding What HappenedFor each eigenvalue:1.Substitute λ into (AλI)(x = 0)2.Solve the resulting system.3.If calculations are correct, there must benonzero solutions.4.The solutions form a line (or subspace) of eigenvectors.If you ever find that only the zero vector works, it means something went wrongbecauseeigenvalues must produce nonzero eigenvectors.Example 2:The General 2×2 MatrixNow consider a general matrix:Let’s explore what we can learn about its eigenvalues.Step 1: Find the Characteristic EquationWe compute:
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Linear Algebra – Eigenvalues and Eigenvectors - Page 10 preview imageStudy GuideThis gives:This is a quadratic equation in λ.Using thequadratic formula:Step 2: Understanding the DiscriminantThe expression under the square root is:This can be rewritten as:Now suppose the matrix issymmetric, meaning:Then the expression becomes:This is the sum of two squares.And a sum of squares is alwaysnonnegative.Therefore:If a matrix is real and symmetric, its eigenvalues are always real numbers.
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