document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164527>Question 1164527</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let Jn be the n × n matrix all of whose entries are 1. Let n > 1 and show that:<BR>\n" );
document.write( "(In − Jn)^−1 = In − 1/(n − 1)Jn </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854346 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2261)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=854346\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To prove that $(I_n - J_n)^{-1} = I_n - \frac{1}{n-1}J_n$, we must show that their product results in the Identity matrix $I_n$. \r<BR>\n" );
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document.write( "### **Key Property of $J_n$**<BR>\n" );
document.write( "Before starting, recall the definition of $J_n$: it is an $n \times n$ matrix where every entry is $1$. <BR>\n" );
document.write( "The product $J_n \cdot J_n$ results in a matrix where each entry is the dot product of a row of $1$s and a column of $1$s. Since there are $n$ entries:<BR>\n" );
document.write( "$$J_n^2 = n J_n$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **The Proof**\r<BR>\n" );
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document.write( "Let $A = (I_n - J_n)$ and $B = (I_n - \frac{1}{n-1}J_n)$. We will compute $A \cdot B$:\r<BR>\n" );
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document.write( "**Step 1: Expand the product**<BR>\n" );
document.write( "$$(I_n - J_n) \left( I_n - \frac{1}{n-1}J_n \right)$$<BR>\n" );
document.write( "Using the distributive property (FOIL):<BR>\n" );
document.write( "$$= I_n \cdot I_n - I_n \cdot \frac{1}{n-1}J_n - J_n \cdot I_n + J_n \cdot \frac{1}{n-1}J_n$$\r<BR>\n" );
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document.write( "**Step 2: Simplify the terms**<BR>\n" );
document.write( "* $I_n \cdot I_n = I_n$<BR>\n" );
document.write( "* $I_n \cdot \frac{1}{n-1}J_n = \frac{1}{n-1}J_n$<BR>\n" );
document.write( "* $J_n \cdot I_n = J_n$<BR>\n" );
document.write( "* $J_n \cdot \frac{1}{n-1}J_n = \frac{1}{n-1}J_n^2$\r<BR>\n" );
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document.write( "Substituting these back:<BR>\n" );
document.write( "$$= I_n - \frac{1}{n-1}J_n - J_n + \frac{1}{n-1}J_n^2$$\r<BR>\n" );
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document.write( "**Step 3: Substitute $J_n^2 = n J_n$**<BR>\n" );
document.write( "$$= I_n - \frac{1}{n-1}J_n - J_n + \frac{n}{n-1}J_n$$\r<BR>\n" );
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document.write( "**Step 4: Combine the $J_n$ terms**<BR>\n" );
document.write( "Factor out $J_n$:<BR>\n" );
document.write( "$$= I_n + \left( -\frac{1}{n-1} - 1 + \frac{n}{n-1} \right) J_n$$\r<BR>\n" );
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document.write( "Find a common denominator for the terms inside the parentheses:<BR>\n" );
document.write( "$$= I_n + \left( \frac{-1 - (n-1) + n}{n-1} \right) J_n$$<BR>\n" );
document.write( "$$= I_n + \left( \frac{-1 - n + 1 + n}{n-1} \right) J_n$$<BR>\n" );
document.write( "$$= I_n + \left( \frac{0}{n-1} \right) J_n$$\r<BR>\n" );
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document.write( "**Step 5: Final Result**<BR>\n" );
document.write( "$$= I_n + 0 = I_n$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Conclusion**<BR>\n" );
document.write( "Since $(I_n - J_n)(I_n - \frac{1}{n-1}J_n) = I_n$, the two matrices are inverses of each other. Therefore:<BR>\n" );
document.write( "$$(I_n - J_n)^{-1} = I_n - \frac{1}{n-1}J_n$$<BR>\n" );
document.write( "$\blacksquare$\r<BR>\n" );
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document.write( "> **Note:** This inverse exists only if $n-1 \neq 0$, which is why the condition $n > 1$ is specified. If $n=1$, the matrix $I_1 - J_1$ becomes $[1] - [1] = [0]$, which is not invertible. </SPAN>\n" );
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