Yankees to 'sign' and honor Matsui on July 28
NEW YORK -- Godzilla is coming back to the Bronx.
The Yankees announced on Thursday that they will celebrate the career of former outfielder Hideki Matsui in a ceremony before their July 28 game against the Rays, honoring No. 55 on what was originally scheduled to be the 55th home game of the season.
Matsui will sign a one-day contract with the Yankees on that day and will officially announce his retirement from baseball as a member of the Yankees. The first 18,000 fans in attendance will receive bobbleheads showing Matsui holding his 2009 World Series MVP trophy.
"I think it's extremely deserving," manager Joe Girardi said. "When you talk about Hideki Matsui, I think about a great player, great teammate. He's someone who was prepared to play every day and gave everything he's got."
The former Yomiuri Giants star played seven seasons for the Yankees from 2003-09, batting .292 with 140 home runs and 597 RBIs during that span. Matsui made the All-Star team in 2003 and '04, and his streak of 518 straight games played remains the longest run to start a career in Major League history.
Matsui played his last game for the Yankees on Nov. 4, 2009, hitting a home run and driving in a World Series record-tying six runs in a Game 6 victory over the Phillies, sealing New York's championship. He was a unanimous pick for Series MVP.
Matsui played three more years after leaving the Yankees following the 2009 season, playing one season each with the Angels and the A's before playing 34 games for the Rays last year. He'll finish his career with a .282 batting average, 175 home runs and 760 RBIs.
"He always found a way to get big hits for the New York Yankees," Girardi said. "You can go back to the World Series and how big he was there. You'll hear every player, to a man, say what a great teammate he was, and I think everyone will look forward to that day."