Linear Algebra Exercises

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Running Head: ALGEBRA 1

Algebra

First Middle Last

Name of Institution

ALGEBRA 2

Task

In a rectangular array of nonnegative real numbers with r rows and c columns, and with r < c,

each column contains at least one positive element.

Prove: There exists a positive element for which the sum of the elements of the intersecting row

(its row sum) is larger than the sum of the elements of the intersecting column (its column sum).

Solution

We will use induction to prove the statement.

Step 1: Proving that the statement holds for the base case, r = 1

Considering r = 1, a single row would have at least two elements. As each column contains at

least one positive element, all the elements would be positive.

Hence, the row sum is bigger than any column (which would be a single element)

Step 2: Proving that the statement holds for r assuming it is correct for any number of

rows less then r. (Note: A row intersecting a column means that they have a mutual positive

element)

Scenario 1: There is a subset of k rows, where the columns set that would intersect any of them

would have a size of < k. For example, row having zero values, or two rows that have single

nonzero elements in the same column.

If such a subset exists, then we can remove out both the k rows and all the intersecting columns.

We will have a new formation of the matrix, with fewer rows and less columns. Therefore, there

will be more columns as compared to rows since more rows than columns have been eliminated.

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1 Running Head: ALGEBRA Algebra First Middle Last Name of Institution ALGEBRA 2 Task In a rectangular array of nonnegative real numbers with r rows and c columns, and with r < c, each column contains at least one positive element. Prove: There exists a positive element for which the sum of the elements of the intersecting row (its row sum) is larger than the sum of the elements of the intersecting column (its column sum). Solution We will use induction to prove the statement. Step 1: Proving that the statement holds for the base case, r = 1 Considering r = 1, a single row would have at least two elements. As each column contains at least one positive element, all the elements would be positive. Hence, the row sum is bigger than any column (which would be a single element) Step 2: Proving that the statement holds for r assuming it is correct for any number of rows less then r. (Note: A row intersecting a column means that they have a mutual positive element) Scenario 1: There is a subset of k rows, where the columns set that would intersect any of them would have a size of < k. For example, row having zero values, or two rows that have single nonzero elements in the same column. If such a subset exists, then we can remove out both the k rows and all the intersecting columns. We will have a new formation of the matrix, with fewer rows and less columns. Therefore, there will be more columns as compared to rows since more rows than columns have been eliminated. ALGEBR ...
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