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Can a square matrix with two identical columns be invertible? Why or why not?

Short Answer

Expert verified
No, a square matrix with two identical columns cannot be invertible as it has linearly dependent columns.

Step by step solution

01

Understanding Invertibility

A square matrix is invertible if there exists another matrix that, when multiplied with the original matrix, yields the identity matrix. This can only happen if the determinant of the matrix is non-zero, which means its columns are linearly independent.
02

Recognizing Column Dependence

If a matrix has two identical columns, these columns are linearly dependent. This means there is a linear relationship where one column can be expressed as a scalar multiple of the other (in this case, the scalar is 1), leading to a zero determinant.
03

Conclusion on Invertibility

Since the presence of two identical columns makes the columns of the matrix linearly dependent, the determinant of the matrix is zero. Therefore, it cannot be invertible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant
The determinant of a matrix is a special value that can be calculated from its elements. It plays a significant role in various aspects of linear algebra. Primarily, it helps to determine if a matrix is invertible or not. For a square matrix, the determinant is calculated using a specific formula that involves all the entries of the matrix.

The determinant is particularly interesting because:
  • If the determinant is zero, the matrix is singular, meaning it does not have an inverse.
  • If the determinant is non-zero, the matrix is non-singular, and an inverse exists.
You can think of the determinant as a scale factor that describes how the volume changes under the transformation defined by the matrix. A zero determinant means any transformation by the matrix squishes the space into a lower dimension, making recovery of the original space via inversion impossible.
Linear Dependence
Linear dependence among the columns of a matrix means that at least one column can be expressed as a linear combination of others. In simpler terms, one column is redundant for expressing the vector space formed by the columns.

Here's how linear dependence affects invertibility:
  • Two identical columns in a matrix imply linear dependence, as one column can be expressed exactly by the other using a scalar of 1.
  • When columns are linearly dependent, it leads to a determinant of zero.
  • A determinant of zero, in turn, implies that the matrix is not invertible.
Linear dependence is crucial for understanding why and how matrices fail to provide the invertibility property. It stems from the fact that you're trying to retrieve a multi-dimensional space with redundant information, which is mathematically infeasible.
Identity Matrix
The identity matrix is a special type of square matrix that acts as the multiplicative identity in the context of matrix multiplication. This means that when any matrix is multiplied by the identity matrix, it remains unchanged. The identity matrix has 1s on its main diagonal and 0s elsewhere.

The role of the identity matrix in invertibility is significant:
  • An invertible matrix, A, has an inverse, denoted as A鈦宦, such that the product A * A鈦宦 equals the identity matrix.
  • This condition is crucial because it means applying the transformation defined by A followed by A鈦宦 brings any vector back to its original form.
  • If a matrix cannot yield the identity matrix when multiplied by another matrix, it is not invertible.
The identity matrix stands as the ultimate goal for verifying invertibility. If a square matrix cannot transform into the identity matrix through multiplication, it falls short of being an invertible entity.

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Most popular questions from this chapter

In Exercises 19鈥24, justify each answer or construction. If the rank of a \(7 \times 6\) matrix \(A\) is \(4,\) what is the dimension of the solution space of \(A \mathbf{x}=\mathbf{0} ?\)

Let \(\mathbf{v}_{1}=\left[\begin{array}{r}{2} \\ {3} \\\ {-5}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{-4} \\ {-5} \\\ {8}\end{array}\right],\) and \(\mathbf{w}=\left[\begin{array}{r}{8} \\ {2} \\\ {-9}\end{array}\right] .\) Determine if \(\mathbf{w}\) is in the subspace of \(\mathbb{R}^{3}\) generated by \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) .

A \(2 \times 200\) data matrix \(D\) contains the coordinates of 200 points. Compute the number of multiplications required to transform these points using two arbitrary \(2 \times 2\) matrices \(A\) and \(B\) . Consider the two possibilities \(A(B D)\) and \((A B) D .\) Discuss the implications of your results for computer graphics calculations.

In Exercises \(3-6,\) the vector \(\mathbf{x}\) is in a subspace \(H\) with a basis \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\} .\) Find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x} .\) $$ \mathbf{b}_{1}=\left[\begin{array}{r}{1} \\ {-3}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{-3} \\ {5}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{-7} \\ {5}\end{array}\right] $$

[M] Some matrix programs, such as MATLAB, have a com- mand to create Hilbert matrices of various sizes. If possible, use an inverse command to compute the inverse of a twelfth- order or larger Hilbert matrix, \(A .\) Compute \(A A^{-1} .\) Report what you find.

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