Problem 8

Question

Let $N$ be a strictly upper truangular $n \times n$ matrix, that is $N=\left(a_{i j}\right)$ and $a_{i j}=0$ if $i \geqq j$. Show that $N^{n}=0$.

Step-by-Step Solution

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Answer

We will use mathematical induction to prove that $N^n=0$, where n is the size of the strictly upper triangular matrix N. The base case for $n=1$ holds, as a $1\times1$ strictly upper triangular matrix is already the zero matrix. Assuming the case holds for any $k\times k$ strictly upper triangular matrix with \(k

1Step 1: Definition of strictly upper triangular matrix

A strictly upper triangular matrix is a square matrix where all the elements on or below the main diagonal are zero, i.e., $a_{ij}=0$ for all $i \geqq j$. In other words, the only non-zero elements are above the diagonal. For example, for a $3 \times 3$ matrix, it would look like: $N = \begin{pmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \end{pmatrix}$

2Step 2: Observations about matrix multiplication

When we multiply two matrices, observe the following pattern in the resulting matrix: Let $N = \begin{pmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \end{pmatrix}$ We multiply N by itself: $N^2=\begin{pmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \end{pmatrix}$ Notice that the resulting matrix will have all elements zero on two diagonals: $N^2=\begin{pmatrix} 0 & 0 & a_{13}a_{23} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

3Step 3: Mathematical induction

Now, we will use mathematical induction to prove that $N^{n}=0$, where n is the size of the strictly upper triangular matrix N. Base case: For $n=1$, a $1\times 1$ strictly upper triangular matrix is already the zero matrix. Hypothesis: Assume that for any $k \times k$ strictly upper triangular matrix $N^{k}=0$, where \(k

Key Concepts

Matrix MultiplicationMathematical InductionThe Zero Matrix

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, an area of mathematics that deals with vectors and matrices. The product of two matrices is calculated by taking the dot product of rows from the first matrix with columns of the second matrix.

In the context of strictly upper triangular matrices, matrix multiplication has a pattern: with each successive multiplication, we see that for a matrix $N^{2}$, $N^{3}$, and continuing onwards, the non-zero elements shift further above the diagonal.

Understanding the Pattern

Let's take a closer look at the multiplication process for a strictly upper triangular matrix. When we multiply such a matrix by itself, each element $a_{ij}$ in the product matrix is calculated as the sum of the products of elements across the corresponding row of the first matrix and down the corresponding column of the second matrix. However, since the elements on or below the diagonal of our strictly upper triangular matrix are zero, many of these product terms are zero, simplifying the calculation.

This characteristic leads to a higher concentration of zero elements in the resulting matrix after each consecutive multiplication, eventually resulting in a matrix of all zeroes, known as the zero matrix.

Mathematical Induction

Mathematical induction is a proof technique that is used to establish the validity of statements for all natural numbers. It consists of two main steps: the base case and the induction step. In mathematical induction, we first show that the statement holds for a starting natural number (the base case). Then, we assume that the statement holds for some natural number $k$, and under this assumption, we prove that the statement holds for the next number $k+1$ (the induction step).

Applying Induction to Upper Triangular Matrices

In the exercise, we use mathematical induction to show that a strictly upper triangular matrix raised to its size power $n$ results in a zero matrix. The base case is trivial for size one. For the induction step, assuming the statement holds for size $k$, we examine the structure of the matrix as if incrementing $k$ to $n$, and show through the multiplication process that $N^n$ must also be the zero matrix. This form of proof is very powerful in mathematics because it allows us to conclude something about an infinite set of cases (all sizes of matrices) based on the validity of the base case and the induction step.

The Zero Matrix

The zero matrix refers to a matrix in which all the elements are zero. In algebra, the zero matrix serves as the identity element for matrix addition; any matrix added to the zero matrix results in the original matrix. It's also analogous to the number zero, in which any number multiplied by zero is zero.

In relation to strictly upper triangular matrices, the significance of the zero matrix comes into play when we discuss powers of matrices. As we see in the exercise, multiplying a strictly upper triangular matrix $N$ by itself $n$ times, where $n$ is the size of the matrix, will result in the zero matrix. This outcome hinges on the fact that the non-zero elements of $N$ are pushed further away from the diagonal with each multiplication, ultimately leaving a matrix full of zeroes - the zero matrix.

Problem 7

Problem 9

Other exercises in this chapter

Problem 6

Let $S$ be a set of $n \times n$ matrices over a field $k$. Show that there exists a column vector $X \neq 0$ of dimension $n$ in $k$, such that \(M

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Problem 7

Let $\mathrm{H}$ be the division ring over the reals generated by elements $i, j, k$ such that $i^{2}=j^{2}=k^{2}=-1$, and $$ i j=-j i=k, \quad j k=-k j=i

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Problem 9

Let $E$ be a vector space over $k$, of dimension $n$. Let $T: E \rightarrow E$ be a linear map such that $T$ is nilpotent, that is $T^{m}=0$ for som

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Problem 10

If $N$ is a nilpotent $n \times n$ matrix, show that $I+N$ is invertible.

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