Suppose that λis an eigenvalue of A. (a) Prove that c λis an eigenvalue of the scalar multiple c

Suppose that λis an eigenvalue of A. (a) Prove that c λis an...

Key Concepts

Matrix Shift by the Identity Matrix

Adding a scalar multiple of the identity matrix to a matrix A results in a uniform shift of all its eigenvalues. For an eigenvalue ? of A with eigenvector v, the matrix A + dI, when applied to v, produces (A + dI)v = Av + d*v. Since Av = ?v, this simplifies to (? + d)v, demonstrating that ? + d is an eigenvalue of A + dI.

Affine Transformation of Eigenvalues

An affine transformation of a matrix combines scalar multiplication and addition of a scalar multiple of the identity matrix. For B = cA + dI, if ? is an eigenvalue of A with eigenvector v, then B*v becomes cAv + d*v. This can be rewritten as (c? + d)v, indicating that each eigenvalue ? of A transforms to c? + d under the mapping B = cA + dI.

Eigenvalues and Eigenvectors

An eigenvalue of a matrix A is a scalar ? such that there exists a nonzero vector v (the eigenvector) satisfying Av = ?v. This concept is fundamental in linear algebra, as it captures how a linear transformation A acts to scale certain directions in the vector space without changing their direction.

Scalar Multiplication of Matrices

When a matrix is multiplied by a scalar c, each eigenvalue of the original matrix A is scaled by c. Specifically, if v is an eigenvector corresponding to the eigenvalue ?, then for the matrix cA, the product cA*v equals c*(Av), which simplifies to c?*v. This shows that c? is an eigenvalue of cA, preserving the eigenvector.

Transcript

00:01 We're asked to prove that if a, b, c, and d are integers, such that a plus b equals c plus d, then it follows.

00:26 The matrix a, which is a, b, c, d, it's a 2x2 matrix, has integer eigenvalues, namely, lambda 1 equals a plus b, and lambda 2 equals a minus c.

00:43 Well, let's find the eigenvalues of this 2x2 matrix.

00:46 To do this, we want to find the determinant of the matrix a minus lambda i, where i is the identity matrix.

00:57 This is the determinant of a minus lambda, b, c, d minus lambda.

01:05 This is the same as a minus lambda times d minus lambda minus bc.

01:13 We want to solve when this is equal to zero, the zeros of the characteristic polynomial.

01:25 So first, all right, this is a quadratic equation.

01:28 We have lambda squared minus a plus d lambda minus bc equals zero.

01:41 And therefore, our eigenvalues are lambda equals, well, first of all, we can actually, using the quadratic formula, find lambda equals a plus d plus or minus the square root of a plus d squared minus four times negative bc all over two this is the same as a plus d plus or minus the square root of well made a mistake actually the characteristic polynomial well this is actually lambda squared minus a plus d plus ad minus bc.

02:52 And so this is actually a plus d squared minus 4 times ad minus bc.

03:08 And so if we distribute foil and do things, we get a squared plus 2ad plus d squared minus 4 ad plus 4bc, all over 2...

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