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Chapter 6: Problem 3

Let \(A\) be a triangular matrix $$ \left(\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1 n} \\ 0 & a_{22} & \ldots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n n} \end{array}\right) $$ Assume that none of the diagonal elements is equal to 0. What is the rank of \(A\) ?

Short Answer

Expert verified
The rank of the given triangular matrix A with non-zero diagonal elements can be found by counting the number of non-zero rows in its row echelon form. Since the matrix is already in the row echelon form and has no zero rows, the rank is equal to the number of rows, which is n. Therefore, Rank(A) = n.

Step by step solution

01

Understanding the given matrix

We are given a triangular matrix (either upper or lower triangular) with non-zero diagonal elements. This means that all elements below (or above) the main diagonal are zeros. We can write the given matrix as: $$ A = \left(\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1 n} \\\ 0 & a_{22} & \ldots & a_{2 n} \\\ \vdots & \vdots & \ddots & \vdots \\\ 0 & 0 & \cdots & a_{n n} \end{array}\right) $$ Furthermore, it is given that \(a_{ii} \neq 0\) for \(i = 1, 2, \ldots, n\).
02

Row Echelon Form

The given matrix is already in the row echelon form. In row echelon form, the first non-zero entry from the left in each row (called the pivot) is to the right of the pivot in the row above it, and all entries below a pivot are zeros.
03

Finding the Rank

The rank of a matrix is defined as the number of non-zero rows in its row echelon form. Since all the diagonal elements are non-zero and there are no zero rows in the given triangular matrix, it has n non-zero rows (or pivots). Therefore, the rank of the given matrix A is n: Rank(A) = n

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Key Concepts

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

Understanding the Triangular Matrix
When we talk about a triangular matrix, we are referring to a square matrix where all the elements above or below the main diagonal are zero. There are two types: upper triangular and lower triangular. An upper triangular matrix has zeros below the diagonal, and a lower triangular one has zeros above it.

The given exercise presents us with a matrix that is already in an upper triangular form, which is half filled with non-zero elements on the diagonal and above, while the other half, below the diagonal, are zeros. This specific format makes it quite straightforward to determine certain properties, such as the determinant and the rank of the matrix.

Picturing this triangular configuration is fundamental for visual learners, as it reinforces the concept that the 'action' of the matrix is focused on the diagonal and the portion immediately above or below it, depending on the type of triangular matrix.
The Row Echelon Form
Every matrix can be transformed into a form known as the row echelon form (REF) through a series of elementary row operations. In REF, each row begins with a leading one (known as a pivot), and all the elements below the pivots are zeros. The pivotal aspect (pun intended) of REF is that it sets the stage for various matrix solutions, including solving linear systems and finding the matrix rank.

What's special about our triangular matrix is that it's inherently in row echelon form, given its zeroes below the diagonal and the condition that all diagonal elements are not zero. This makes our jobs easier since there's no need for additional row manipulations to reach REF. This property is a freebie - it saves us time and effort in our calculations. For the students who've tangled with matrices that require row-reduction to reach this form, recognizing a matrix already in REF is a little victory worth celebrating!
Importance of Non-Zero Diagonal Elements
The presence of non-zero diagonal elements is a crucial aspect of our triangular matrix. Why? Well, let's think about the game of connect-the-dots. If any of the dots (in our case, the diagonal elements) were missing, we couldn't complete the picture (here, the matrix's foundational structure).

In a triangular matrix, these non-zero diagonal elements serve as the 'anchors' that hold the entire matrix together. They are integral in determining the rank of the matrix and ensuring that it functions appropriately in mathematical operations. Essentially, it's these non-zero elements that confirm the matrix's full rank of n, where n is the number of rows or columns (since all our matrices are square in this context).

It's the guarantee of these non-missing 'dots' that ensures the strength and stability of the matrix, allowing it to have a clear and defined rank. For students grappling with the concept of rank, associating it with this visual image of a complete, robust structure might just make the 'aha' moment stick.

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Most popular questions from this chapter

(a) Let \(V\) be the vector space of all \(n \times n\) matrices over \(\mathbf{R}\), and define the scalar product of two matrices \(A, B\) by $$ \langle A, B\rangle=\operatorname{tr}(A B) $$ where tr is the trace (sum of the diagonal elements). Show that this is a scalar product and that it is non-degenerate. (b) If \(A\) is a real symmetric matrix, show that \(\operatorname{tr}(A A) \geqq 0\), and \(\operatorname{tr}(A A)\) \(>0\) if \(A \neq 0 .\) Thus the trace defines a positive definite scalar product on the space of real symmetric matrices. (c) Let \(V\) be the vector space of real \(n \times n\) symmetrie matrices. What is \(\operatorname{dim} V ?\) What is the dimension of the subspace \(W\) consisting of those matrices \(A\) such that \(\operatorname{tr}(A)=0 ?\) What is the dimension of the orthogonal complement \(W^{\perp}\) relative to the positive definite sealar product of part (b)?

Find the dimension of the space of solutions of the following systems of equations. Also find a basis for this space of solutions. (a) \(2 x+y-z=0\) (b) \(x-y+z=0\) \(y+z=0\) (c) \(4 x+7 y-\pi z=0\) (d) \(x+y+z=0\) \(2 x-y+z=0 \quad x-y=0\) \(y+z=0\)

Let \(A X=B\) be a system of linear equations, where \(A\) is an \(m \times n\) matrix, \(X\) is an \(n\) -vector, and \(B\) is an \(m\) -vector. Assume that there is one solution \(X=X_{0} .\) Show that every solution is of the form \(X_{0}+Y\), where \(Y\) is a solution of the homogeneous system \(A Y=O\), and conversely any vector of the form \(X_{0}+Y\) is a solution.

Let \(W\) be the subspace of \(\mathbf{C}^{3}\) generated by the vector \((1, i, 0)\). Find a basis of \(W \perp\) in \(C^{3}\) (with respect to the ordinary dot produet of vectors).

Let \(V\) be a vector space of dimension \(n\) over the field \(K .\) Let \(\psi, \varphi\) be two non-zero functionals on \(V\). Assume that there is no element \(c \in K, c \neq 0\) such that \(\psi=c \varphi .\) Show that $$ (\text { Ker } \varphi) \cap(\operatorname{Ker} \psi) $$ has dimension \(n-2\).
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