Question

Let $A \in M_{nn}(\mathbb{C})$ be a matrix with $i, j$-th entry $a_{ij} \in \mathbb{C}$. If $|a_{ii}| > \sum_{j \neq i} |a_{ij}|$ for all $1 \leq i \leq n$, then show that A is invertible. 2) Let V be an n-dimensional vector space over F. Let S be finite subset of V such that $|S| \geq n+1$. $|S| > n+1$. Show that there exists a non-zero function f: S \to F such that $\sum_{v \in S} f(v)v = 0$ and $\sum_{v \in S} f(v) = 0

          Let $A \in M_{nn}(\mathbb{C})$ be a matrix with $i, j$-th
entry $a_{ij} \in \mathbb{C}$. If $|a_{ii}| > \sum_{j \neq i} |a_{ij}|$ for all
$1 \leq i \leq n$, then show that A is invertible.
2) Let V be an n-dimensional vector space
over F. Let S be finite subset of
V such that $|S| \geq n+1$. $|S| > n+1$.
Show that there exists a non-zero function
f: S \to F such that
$\sum_{v \in S} f(v)v = 0$ and $\sum_{v \in S} f(v) = 0

Let A ∈ Mnn(ℂ) be a matrix with i, j-th
entry aij∈ℂ. If |aii| > ∑j ≠ i |aij| for all
1 ≤ i ≤ n, then show that A is invertible.
2) Let V be an n-dimensional vector space
over F. Let S be finite subset of
V such that |S| ≥ n+1. |S| > n+1.
Show that there exists a non-zero function
f: S →F such that
∑v ∈ S f(v)v = 0 and ∑v ∈ S f(v) = 0

Added by Tracy C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Manisha Sarker

05/09/2021

Step 1

We will use the fact that the determinant of a matrix is equal to the sum of the products of the elements of any row or column, each multiplied by its cofactor. Show more…

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Let $A \in M_{nn}(\mathbb{C})$ be a matrix with $i,j$-th entry $a_{ij} \in \mathbb{C}$. If $|a_{ii}| > \sum_{j \neq i} |a_{ij}|$, for all $1 \leq i \leq n$, then show that $A$ is invertible. 2) Let $V$ be an $n$-dimensional vector space over $F$. Let $S$ be finite subset of $V$ such that $|S| \geq n+1$ $|S| > n+1$. Show that there exists a non-zero function $f: S \rightarrow F$ such that $\sum_{v \in S} f(v)v = 0$ and $\sum_{v \in S} f(v) = 0$