Answer :
Answer:
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 [tex]3 \times 3[/tex] diagonal matrices that is,
[tex]\left[\begin{array}{ccc}0&1&0\\0&0&6\\0&0&0\end{array}\right][/tex]
here [tex]A^3[/tex]= 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 [tex]A =\left[\begin{array}{ccc}0&1\\0&0\end{array}\right][/tex] 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 [tex]A= PDP^{-1}[/tex] 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 [tex]R^n[/tex]. A has eigen vector and A is [tex]n \times n[/tex] matrix, the number of A eigen values = given vector of A, that's why it is correct.
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 [tex]A=PDP^{-1[/tex] 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,
[tex]A=\left[\begin{array}{ccc}0&2&0\\0&0&1\\0&0&0\end{array}\right][/tex]
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 [tex]A=\left[\begin{array}{cc}0&0\\0&0\end{array}\right][/tex]
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 [tex]A = PDP^{-1}[/tex] 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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