A, B, P, and D are n ×n matrices. Mark each statement True...

$A, B, P,$ and $D$ are $n \times n$ matrices. Mark each statement True or False. Justify each answer. (Study Theorems 5 and 6 and the examples in this section carefully before you try these exercises.) a. $A$ is diagonalizable if $A=P D P^{-1}$ for some matrix $D$ and some invertible matrix $P$ . b. If $\mathbb{R}^{n}$ has a basis of eigenvectors of $A,$ then $A$ is diagonalizable. c. $A$ is diagonalizable if and only if $A$ has $n$ eigenvalues, counting multiplicies. d. If $A$ is diagonalizable, then $A$ is invertible.

$A, B, P,$ and $D$ are $n \times n$ matrices. Mark each statement True or False. Justify each answer. (Study Theorems 5 and 6 and the examples in this section carefully before you try these exercises.)
a. $A$ is diagonalizable if $A=P D P^{-1}$ for some matrix $D$ and some invertible matrix $P$ .
b. If $\mathbb{R}^{n}$ has a basis of eigenvectors of $A,$ then $A$ is diagonalizable.
c. $A$ is diagonalizable if and only if $A$ has $n$ eigenvalues, counting multiplicies.
d. If $A$ is diagonalizable, then $A$ is invertible.

Key Concepts

Matrix Invertibility

Matrix invertibility is a property signifying that a matrix has an inverse. It is closely related to the eigenvalues of the matrix; a matrix is invertible if and only if it does not have 0 as an eigenvalue. This means that even if a matrix is diagonalizable, it might still be non-invertible if one of its eigenvalues is zero.

Diagonalizability

Diagonalizability is the property of a matrix where it can be expressed in the form PDP?¹, with D being a diagonal matrix and P invertible. This representation simplifies many matrix computations, as the diagonal matrix D contains the eigenvalues of the original matrix and the columns of P correspond to its eigenvectors.

Eigenvector Basis

An eigenvector basis is a set of linearly independent eigenvectors that spans the entire vector space. If a matrix has such a basis, then every vector in the space can be written as a linear combination of eigenvectors, which is equivalent to being able to diagonalize the matrix. This concept is central in determining diagonalizability.

Multiplicity of Eigenvalues

The multiplicity of eigenvalues involves both algebraic multiplicity (the number of times an eigenvalue appears as a root of the characteristic polynomial) and geometric multiplicity (the dimension of the corresponding eigenspace). A matrix is diagonalizable if for every eigenvalue its geometric multiplicity equals its algebraic multiplicity, ensuring enough eigenvectors for diagonalization.

Transcript

00:01 Okay, so problem 21 is a series of true or false.

00:05 First statement, if we can write a equals a times a matrix d times p minus 1 for p an invertible matrix, then a is diagonalizable.

00:19 The first statement is false because a is only diagonalizable if d is a diagonalized matrix.

00:32 For a to be, not big a, writing too much matrices this morning.

00:41 For a to be true, so for this placement to be true, d, so matrix d must be diagonal, a diagonal matrix.

01:02 As the statement is written right now, so d is any matrix.

01:09 The statement is false, a diagonalize.

01:11 Matrix can be written as p and invertible matrix times d a diagonal matrix times p minus 1.

01:19 Statement 2 if the rn has a basis of eigenvector of a that means a is diagonalizable and that statement is true because a basis of rn has an linearly independent vectors and and a is diagonalizable if it is not if it has an linearly independent pagan vectors and since those agen vectors are a basis of n they are a basis of n they are linearly independent, therefore a is diagonalizable.

02:44 If a has n hegan values counting multiplicity, then a is diagonalizable.

02:52 That statement is false because let's take, so every matrix has n hegan values counting multiplicity.

03:07 So for the statement to be true, we should write a as n distinct hagen values, but n -agin values counting multiplicity is always true.

03:19 If you need an example, we can look at problem 17 from this section where a is equal to 4 -0 -1 -4 -0 -0 -05.

03:34 So the agon values are lambda 1 equals 4 with a multiplicity of 2.

03:41 Let's write it properly with a multiplicity of 2 and lambda 2 equals 5 with a multiplicity of 1.

03:54 So 2 plus 1 equals 3.

03:57 So that you have n eigenvalues counting multiplicity, but it only has two linearly independent eigenvectors, which are 010.

04:07 And these are the egg inventors as zero one.

04:15 So only tooling or the independent vector and a is not diagonalizable.

04:21 If you need more details, i encourage you to go look at the solution of problem 17.

04:28 There's also video for that...

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