Question

If \(u\) is a number and \(A\) is an \(n \times n\) matrix, prove that \(\operatorname{det}(u A)=u^{n}\) det \(A\) by induction on \(n,\) using only the definition of \(\operatorname{det} A\).

Short answer

The formula \(\operatorname{det}(u A)=u^{n}\operatorname{det} A\) holds for all \(n\times n\) matrices by induction.

Step-by-step solution

- Base Case

First, consider the base case where the matrix \(A\) is a \(1 \times 1\) matrix. Let \(A = [a_{11}]\), so \(\operatorname{det}(A) = a_{11}\). For the matrix \(uA = [u a_{11}]\), the determinant is \(\operatorname{det}(uA) = u a_{11}\). Thus, \(u^1 \operatorname{det}(A) = u a_{11}\). The base case holds true as \(\operatorname{det}(uA) = u \operatorname{det}(A)\).

- Inductive Hypothesis

Assume that for an \(n \times n\) matrix, the statement holds true: \(\operatorname{det}(uA) = u^n \operatorname{det}(A)\). This is our inductive hypothesis.

- Inductive Step

Consider an \((n+1) \times (n+1)\) matrix \(B\). By expanding the determinant of \(B\) along the first row, we can express \(\operatorname{det}(B)\) as a sum of terms involving \(n \times n\) minors as follows:\[ \operatorname{det}(B) = \sum_{j=1}^{n+1} (-1)^{1+j} b_{1j} M_{1j} \]where \(M_{1j}\) is the \(n \times n\) minor obtained by removing the first row and \(j\)-th column of \(B\).

- Apply Inductive Hypothesis to Minors

Now consider \(uB\), which is the matrix where each element of \(B\) is multiplied by \(u\). The determinant is:\[ \operatorname{det}(uB) = \sum_{j=1}^{n+1} (-1)^{1+j} (u b_{1j}) \operatorname{det}(u M_{1j}) \]By the inductive hypothesis, we have \(\operatorname{det}(uM_{1j}) = u^n \operatorname{det}(M_{1j})\). So,\[ \operatorname{det}(uB) = u \sum_{j=1}^{n+1} (-1)^{1+j} b_{1j} u^n \operatorname{det}(M_{1j}) = u^{n+1} \sum_{j=1}^{n+1} (-1)^{1+j} b_{1j} \operatorname{det}(M_{1j}) \]Thus, \( \operatorname{det}(uB) = u^{n+1} \operatorname{det}(B) \).

- Conclusion

Since the base case holds true and the inductive step has confirmed the statement for an \((n+1) \times (n+1)\) matrix, by mathematical induction the formula \(\operatorname{det}(uA) = u^n \operatorname{det}(A)\) holds for all \(n \times n\) matrices.

Key concepts

mathematical induction

Mathematical induction is a powerful proof technique used to verify a statement for all natural numbers. It works similarly to dominoes falling in a line. If the first domino falls and every domino knocks over the next one, all will eventually fall. Here's how it works in the context of proving a determinant property using induction:

• Start with the base case. Prove that the statement holds for the smallest natural number, which is usually 1. In our case, we proved that for a 1x1 matrix, \(\operatorname{det}(uA) = u \, \operatorname{det}(A)\).
• Assume true for an arbitrary size. This is the inductive hypothesis. Assume that for an \(n \times n\) matrix, \(\operatorname{det}(uA) = u^n \, \operatorname{det}(A)\).
• Prove for the next size. Demonstrate that if the statement holds for \(n\), then it must hold for \(n+1\). Applying this step ensures that our formula \(\operatorname{det}(uA) = u^n \, \operatorname{det}(A)\) works for any sized \(n \, \times \, n\) matrix. This step was key in showing the formula applies universally by transforming an \(n+1 \times n+1\) matrix using the inductive hypothesis.

By ensuring both the base case and the inductive step work, string the argument together fully covers all positive integers.

matrix multiplication

Matrix multiplication is essential in handling determinants, especially when working with scaled matrices like \(uA\). It might seem complicated at first, but here's how it works:

• Formula is different. Unlike regular multiplication, matrix multiplication requires the row elements of the first matrix to multiply with the column elements of the second matrix.
• Size matters. You can only multiply two matrices if the number of columns in the first equals the number of rows in the second.
• Matrix scaling. When a matrix is multiplied by a scalar, like \(u\), it scales each element. For a matrix \(A\), the matrix \(uA\) has all its elements multiplied by \(u\). This affects the determinant significantly, which is why it links to the property we want to prove: \(\operatorname{det}(uA) = u^n \, \operatorname{det}(A)\).

Understanding how matrix elements combine or scale provides insight into how determinants react to scalar multiplication.

determinant properties

The determinants of matrices are crucial in linear algebra, known for their ability to summarize a matrix's properties into a single value. Here's what you need to know about determinant properties in this context:

• Basic definition. The determinant of a matrix, \(A\), often denoted as \(\operatorname{det}(A)\), is a function that maps a square matrix to a single scalar. For a 2x2 matrix, it's calculated as \(ad-bc\) for elements \([a, b; c, d]\).
• Multiplicative with scalars. One key property is that multiplying every element in a matrix by a scalar \(u\) scales the determinant by \(u^n\), where \(n\) is the size of the matrix.
• Importance. The determinant is zero if a matrix is singular (non-invertible), indicating linear dependence among rows/columns. This property is vital if you calculate or change the determinant during manipulations, including when scaling a matrix.

Determinants transform when matrices scale or multiply, connecting back to why \(\operatorname{det}(uA)\) equals \(u^n \, \operatorname{det}(A)\) forms an important aspect of these calculations.