The Linear Dependence of the Columns in Matrix A^25 | A Counterexample

If A is a 4×4 matrix, then the columns of A^25 are linearly independent. (T/F)

The statement is true

The statement is true.

To understand why, let’s break down the question step by step.

First, let’s define what it means for a set of vectors (in this case, the columns of a matrix) to be linearly independent. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. In other words, if we have vectors v1, v2, …, vn, they are linearly independent if and only if the equation a1v1 + a2v2 + … + anvn = 0, where a1, a2, …, an are scalars, only has the trivial solution a1 = a2 = … = an = 0.

Now, let’s consider the matrix A^25. This notation means that we are raising matrix A to the power of 25, which is equivalent to multiplying A by itself 25 times. In other words, A^25 = AAA…A (25 times).

Since A is a 4×4 matrix, it has 4 columns. Let’s label them as c1, c2, c3, and c4. The question asks if the columns of A^25 are linearly independent, so this means we want to determine if the columns of A^25 (let’s call them d1, d2, d3, and d4) satisfy the condition that a1d1 + a2d2 + a3d3 + a4d4 = 0 only has the trivial solution a1 = a2 = a3 = a4 = 0.

To answer this question, we need to consider the relationship between A and A^25. When we raise a square matrix to a positive integer power, the columns of the resulting matrix are still linearly dependent. This is because matrix multiplication is essentially a series of linear combinations. Each column of A^25 is obtained by multiplying A by itself multiple times and combining the columns of the intermediate results.

Therefore, the columns of A^25 still retain the linear dependence of the original matrix A. Since the original matrix A has 4 columns, the columns of A^25 are linearly dependent as well.

Hence, the statement that the columns of A^25 are linearly independent is false.

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