Mathematics, 28.05.2020 16:58 adrian08022

(1 point) A, P and D are n×n matrices. Check the true statements below: A. A is diagonalizable if and only if A has n eigenvalues, counting multiplicities. B. If A is diagonalizable, then A is invertible. C. A is diagonalizable if A=PDP−1 for some diagonal matrix D and some invertible matrix P. D. If there exists a basis for Rn consisting entirely of eigenvectors of A, then A is diagonalizable.

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Answer from: robyn88

A) A is diagonalizable if and only if A has n eigenvalues, counting multiplicities. False.

B). If A is diagonalizable, then A is invertible. False.

C). A is diagonalizable if $A=PDP^{-1$ for some diagonal matrix D and some invertible matrix P. True.

D) If there exists a basis for Rn consisting entirely of eigenvectors of A, then A is diagonalizable. True.

We have A, P and D are n×n matrices.

Now, In option A,

Let A be diagonal matrices that is,

$A=\left[\begin{array}{ccc}0&2&0\\0&0&1\\0&0&0\end{array}\right]$

Here the eigen values is 0 with multiplicity 3.

In this A is the nilpotent matrix and A has three values that are equal to 0 but it is not a diagonalizable matrix. (False).

In option B,

Consider the matrix $A=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

It the 2×2 zero matrix. The zero matrix is a diagonal matrix, and thus it is diagonalizable. However, the zero matrix is not invertible as its determinant is zero.

|A| =0 implies that A is not invertible. ( False)

In option C,

A square matrix A is diagonalizable if A is similar to a diagonal matrix.

This means $A = PDP^{-1}$ for some invertible P and diagonal D, with all matrices being n × n. (True).

In option D

An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. A basis of Rn of eigenvectors consists of n linearly independent eigen vectors. (True).

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Answer from: ErrorNameTaken505

Following are the answers to this questions:

A) False.

B) False.

C) True.

D) True.

Step-by-step explanation:

In option A,

Let A is $3 \times 3$ diagonal matrices that is,

$\left[\begin{array}{ccc}0&1&0\\0&0&6\\0&0&0\end{array}\right]$

here A^3 = 0, In this A is the nilpotent matrix and A has three values that are equal to 0 but it is not a diagonalizable matrix.

In option B,

If $A =\left[\begin{array}{ccc}0&1\\0&0\end{array}\right]$ in given value A elements are 0 and 1 as a distinct eigen value, that's why A is diagonalizable but |A| =0 its means A is not invertible, that's why it is false.

In option C,

It is given that $A= PDP^{-1}$ where A~ D, which is A is similar to diagonal matrices, that's why it is true.

In option D,

It is given that A has eigen vector, which leases from R^n . A has eigen vector and A is $n \times n$ matrix, the number of A eigen values = given vector of A, that's why it is correct.